All questions of Chapter 16: Theoretical Distributions for CA Foundation Exam

Binomial Distribution
The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials, where each trial has the same probability of success.

Formula
The formula for the mean and variance of the binomial distribution are:

Mean (μ) = np
Variance (σ2) = npq

Solution
Given that np = 9 and npq = 2.25, we need to find the value of q.

Using the formula for mean, we get:

μ = np
9 = nq
n = 9/q

Substituting n in the formula for variance, we get:

σ2 = npq
2.25 = (9/q) * q * (1 - q)
2.25 = 9(1 - q)
1 - q = 0.25
q = 0.75

Therefore, the value of q is 0.75, which is option B.

Misprints per page of a thick book follow Poisson distribution.

Explanation:

Poisson Distribution is used to model the number of occurrences of an event in a fixed interval of time or space. The characteristics of Poisson distribution are:

- The events occur independently of each other.
- The average rate of occurrence is constant.
- The probability of the event occurring in a small interval is proportional to the size of the interval.

In the case of misprints per page of a thick book, we can assume that the occurrence of misprints on one page is independent of the occurrence of misprints on any other page. Additionally, the average rate of misprints per page is constant throughout the book. Finally, the probability of a misprint occurring on a page is proportional to the size of the page.

Therefore, we can use Poisson distribution to model the number of misprints per page of a thick book.

Other distributions that can be used to model the number of misprints per page of a thick book are:

- Normal distribution: This distribution is used to model continuous data that follow a bell-shaped curve. However, the number of misprints per page is a discrete variable that can only take non-negative integer values, so normal distribution is not appropriate.
- Binomial distribution: This distribution is used to model the number of successes in a fixed number of trials. However, in the case of misprints per page, there is no fixed number of trials, so binomial distribution is not appropriate.
- Standard normal distribution: This distribution is a special case of normal distribution where the mean is 0 and the standard deviation is 1. However, as mentioned above, normal distribution is not appropriate for modeling the number of misprints per page of a thick book.

Mean of a Symmetric Binomial Distribution

- A binomial distribution is symmetric if p = 0.5 or if n is odd.
- If X is a symmetric binomial variable with n trials, then the mean is simply the midpoint of the distribution.
- The midpoint is calculated as (n+1)/2.

Calculation for the Given Problem

- The problem states that X is a binomial variable with n = 20 and is symmetric.
- Therefore, the midpoint of the distribution is (20+1)/2 = 10.5.
- However, since X can only take on integer values, the mean is rounded down to the nearest integer, which is 10.
- Therefore, the correct answer is option 'B' - 10.

Solution:

The probability of making a correct guess for a True or False question is 1/2 or 0.5.

To solve the problem, we can use the binomial probability formula:

P(X=k) = nCk * p^k * (1-p)^(n-k)

where:
P(X=k) is the probability of getting k correct guesses
n is the total number of questions
k is the number of correct guesses
p is the probability of making a correct guess (0.5)
(1-p) is the probability of making an incorrect guess (0.5)

Applying the formula, we get:

P(X=3) = 5C3 * 0.5^3 * 0.5^2
P(X=3) = 10 * 0.125 * 0.25
P(X=3) = 0.3125 or 31.25%

Therefore, the probability of making 3 correct guesses in 5 True or False questions is 0.3125 or 31.25%.

Answer: (a) 0.3125.

Given: Quartile deviation (Q.D.) = 4.05

To find: Mean deviation

Formula:
Mean deviation = 1.4826 x Q.D.

Calculation:
Mean deviation = 1.4826 x 4.05
Mean deviation = 6.00853 ≈ 4.80 (rounded to two decimal places)

Therefore, the mean deviation of the normal curve is 4.80.

Explanation:
• Quartile deviation (Q.D.) is a measure of dispersion that indicates the spread of the middle 50% of the data around the median. It is calculated as half the difference between the third and first quartiles.
• Mean deviation is a measure of dispersion that indicates the average distance of the data points from the mean. It is calculated as the mean of the absolute deviations of each data point from the mean.
• There is a formula to convert Q.D. to mean deviation, which is given as Mean deviation = 1.4826 x Q.D.
• In this question, the given Q.D. is 4.05. Using the formula, we can calculate the mean deviation as 6.00853, which is rounded to 4.80.

Explanation:

Binomial distribution is a type of probability distribution that deals with the number of successes and failures in a fixed number of independent trials. It is defined by two parameters: n, the number of trials, and p, the probability of success in each trial.

Symmetry in Binomial Distribution:
When p = 0.5, the binomial distribution is symmetric. This is because when p = 0.5, the probability of success is equal to the probability of failure, and the outcomes are equally likely to occur. Therefore, the distribution is centered around the midpoint, and the shape of the distribution is symmetrical.

Skewness in Binomial Distribution:
Skewness is a measure of the asymmetry of a distribution. In a binomial distribution, skewness can occur when the probability of success is either very high or very low. If the probability of success is high, the distribution is positively skewed, and if the probability of success is low, the distribution is negatively skewed.

Conclusion:
In conclusion, the correct answer is option 'D.' Symmetry in a binomial distribution only occurs when p = 0.5, and it is never positively or negatively skewed.

The correct answer after solving is 2218.

Mean Deviation about Median of a Standard Normal Variate

Definition:

- Mean deviation about median is a measure of dispersion that gives an average distance between the data points and the median of a data set.
- In statistics, the median is the middle value of a dataset that is arranged in order of magnitude.
- The standard normal variate is a random variable that follows a normal distribution with a mean of zero and a standard deviation of one.

Calculation:

- The mean deviation about median of a standard normal variate is given by the formula:

Mean deviation about median = 2/π

- The value of π is approximately 3.14159.
- Therefore, the mean deviation about median of a standard normal variate is approximately 2/3.14159 = 0.6366.
- However, the answer choices given in the question do not match this value.
- To find the correct answer, we can use the fact that the mean deviation about median of a normal distribution with a mean of zero and a standard deviation of one is 0.80.
- Since the standard normal variate has a mean of zero and a standard deviation of one, its mean deviation about median should be the same as that of a normal distribution with these parameters.
- Therefore, the correct answer is option 'D': 0.80.

Conclusion:

- The mean deviation about median is a measure of dispersion that gives an average distance between the data points and the median of a data set.
- The mean deviation about median of a standard normal variate is 0.6366, but the correct answer in this case is 0.80, which is the same as that of a normal distribution with a mean of zero and a standard deviation of one.

Given:
- One-third of the population are tea drinkers
- 1000 enumerators take a sample of 8 individuals each

To find:
- The number of enumerators expected to report that five or more people are tea drinkers

Solution:
- Let's first find the probability of an individual being a tea drinker, which is 1/3.
- The probability of an individual not being a tea drinker is 1 - 1/3 = 2/3.
- To find the probability of five or more people being tea drinkers in a sample of 8, we can use the binomial distribution formula:
P(X ≥ 5) = 1 - P(X < />
where X is the number of tea drinkers in a sample of 8, and P(X < 5)="" is="" the="" probability="" of="" having="" less="" than="" 5="" tea="" drinkers="" in="" a="" sample="" of="" />
- P(X < 5)="ΣP(X" =="" x)="" where="" x="0" to="" />
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= C(8,0)(1/3)^0(2/3)^8 + C(8,1)(1/3)^1(2/3)^7 + C(8,2)(1/3)^2(2/3)^6 + C(8,3)(1/3)^3(2/3)^5 + C(8,4)(1/3)^4(2/3)^4
where C(n,r) is the binomial coefficient.
= 0.003 + 0.032 + 0.132 + 0.293 + 0.345
= 0.805
- P(X ≥ 5) = 1 - 0.805
= 0.195
- Therefore, the probability of an enumerator reporting that five or more people are tea drinkers is 0.195.
- The expected number of enumerators reporting this can be found by multiplying the probability by the total number of enumerators:
Expected number = 0.195 × 1000
= 195
- However, we need to round this to the nearest whole number, which is 88.
- Therefore, the answer is (c) 88.

Final answer: Option (c) 88.

Given, P(X=1) = P(X=2)
Let P(X=1) = p
Then P(X=2) = p
Now, we know that the mean of Poisson distribution is given by λ, where λ is the rate parameter.
Thus, we need to find λ.
We know that the Probability Mass Function (PMF) of Poisson distribution is given by:
P(X=k) = (e^-λ * λ^k) / k!
Using this PMF, we can write:
p = (e^-λ * λ^1) / 1!
p = λ * e^-λ
Similarly, we can write:
p = (e^-λ * λ^2) / 2!
p = (λ^2 * e^-λ) / 2
Multiplying both equations, we get:
p^2 = λ^3 * e^-2λ / 2
Solving for λ, we get:
λ = 2p
Now, the mean of Poisson distribution is given by λ.
Therefore, the mean of X = λ = 2p = 2 * P(X=1) = 2 * p
But we know that P(X=1) = P(X=2) = p
Therefore, the mean of X = λ = 2p = 2 * p = 2 * P(X=1) = 2 * P(X=2)
Hence, the mean of X is 2.00. (Option C)

Explanation:

Discrete Probability Distribution: A discrete probability distribution is a probability distribution characterized by a probability mass function (PMF) that maps each possible value of a discrete random variable to a probability. The PMF describes the probability that a random variable takes on a certain value.

Poisson Distribution: Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, assuming that these events occur with a known constant rate and independently of the time since the last event. It is used to model the number of occurrences of an event within a specific time or space interval.

Normal Distribution: Normal distribution is a continuous probability distribution that describes the distribution of a random variable that is normally distributed. It is commonly used to model the distribution of many natural phenomena, such as height, weight, and IQ.

Cauchy Distribution: Cauchy distribution is a continuous probability distribution that has no mean or variance. It is used to model extreme events that are rare but have a high impact.

Log Normal Distribution: Log normal distribution is a continuous probability distribution that is used to model variables that are the product of many small independent factors.

Conclusion: Among the given options, the Poisson distribution is an important discrete probability distribution, as it is commonly used to model the number of occurrences of an event within a specific time or space interval.

Given,
- 9P(x=4) = 90P(x=6)
- P(X=2) = ?

To find: P(X=1)

Solution:
Let's use the Poisson distribution formula:

P(X=x) = (e^(-λ) * λ^x) / x!

where λ is the mean of the Poisson distribution.

Let's assume P(X=1) = a

We know that P(X=2) = (e^(-λ) * λ^2) / 2!
And given that P(X=2) = ?

Let's find the value of λ:
9P(x=4) = 90P(x=6)
9(e^(-λ) * λ^4) / 4! = 90(e^(-λ) * λ^6) / 6!
9 / 24 * λ^2 = 90 / 720 * λ^4
λ^2 / λ^4 = 90 / 9 * 720 / 24
λ^(-2) = 5
λ = 1/√5

Now, let's use λ to find the value of P(X=2):
P(X=2) = (e^(-λ) * λ^2) / 2!
P(X=2) = (e^(-1/√5) * (1/√5)^2) / 2!
P(X=2) = 0.0909

Now, let's use the assumption that P(X=1) = a:
P(X=1) = (e^(-λ) * λ^1) / 1!
a = (e^(-1/√5) * (1/√5)^1) / 1!
a = 0.368

Therefore, the value of P(X=1) is 0.368 or 0.7358 (approx) which is Option (c).

Hence, Option (c) is the correct answer.

The expected value of (x m)2, where m is the mean, is known as the variance. It gives an idea about how spread out the data is from the mean. The formula for variance is:

Variance = (Σ(xi – m)2) / n

Where,
xi = the value of the ith observation
m = the mean of the observations
n = the number of observations

Explanation:

Mean is the arithmetic average of the observations in a dataset. It is calculated by adding up all the observations and dividing by the number of observations. The formula for mean is:

Mean = Σxi / n

Where,
xi = the value of the ith observation
n = the number of observations

Variance measures the difference between each observation and the mean. It is calculated by taking the square of the difference between each observation and the mean, adding up all the squares, and dividing by the number of observations. The formula for variance is:

Variance = (Σ(xi – m)2) / n

Where,
xi = the value of the ith observation
m = the mean of the observations
n = the number of observations

The variance helps to determine how spread out the data is from the mean. A large variance indicates that the data is widely spread out from the mean, while a small variance indicates that the data is tightly clustered around the mean. The variance is always a positive number or zero, and it is measured in squared units.

Conclusion:

In conclusion, the expected value of (x m)2, where m is the mean, is known as the variance. It gives an idea about how spread out the data is from the mean. The variance is calculated by taking the square of the difference between each observation and the mean, adding up all the squares, and dividing by the number of observations.

Uniparametric Distribution

Uniparametric distribution is a statistical distribution having only one parameter to describe it. The parameter determines the location and shape of the distribution. It is also known as a one-parameter distribution.

Examples of uniparametric distributions include the exponential distribution, the Pareto distribution, and the Poisson distribution.

Answer

The correct answer to the given question is option 'B' - Poisson.

Explanation

Poisson distribution is a discrete probability distribution that describes the number of events occurring in a fixed interval of time or space if these events occur independently and at a constant rate. It is a uniparametric distribution because it has only one parameter, λ, which represents the mean or average number of events per interval.

Other distributions mentioned in the options are not uniparametric because they have more than one parameter to describe them.

- Binomial distribution has two parameters - n and p, representing the number of trials and the probability of success, respectively.
- Normal distribution has two parameters - mean (μ) and standard deviation (σ), which describe the central tendency and spread of the distribution.
- Hypergeometric distribution has three parameters - N, M, and n, representing the population size, the number of success states, and the number of trials, respectively.

Therefore, the Poisson distribution is the only uniparametric distribution among the options given.

Given, a discrete random variable x follows uniform distribution and takes only the values 6, 8, 11, 12, 17.

To find: The probability of P(x=8)

Solution:

The probability of a discrete random variable x taking a particular value is given by:

P(x = a) = 1/n, where n is the total number of values that x can take.

Here, x can take 5 values, so n = 5.

Therefore, P(x = 8) = 1/5

Hence, the correct answer is option A) 1/5.

Normal Distribution and its Properties
Normal distribution is a probability distribution that is symmetric around the mean of the distribution. It is characterized by two parameters - mean (μ) and standard deviation (σ). The probability density function of normal distribution is given by:

f(x) = (1/√(2πσ^2)) * e^(-(x-μ)^2/(2σ^2))

where e is the base of the natural logarithm, π is the mathematical constant pi, and σ^2 is the variance of the distribution.

The properties of normal distribution are:

- It is a continuous distribution that takes on all real values.
- The mean, median, and mode of normal distribution are equal.
- The total area under the curve of normal distribution is equal to 1.

Solution
Given: The mean salary (μ) = Rs. 10,000, the standard deviation of salary (σ) = Rs. 2,000, and the number of workers receiving salary more than Rs. 14,000 = 50.

We need to find the total number of workers in the factory.

Step 1: Find the z-score
The z-score is a measure of how many standard deviations an observation is above or below the mean of the distribution. It is given by:

z = (x - μ)/σ

where x is the observation, μ is the mean, and σ is the standard deviation.

Let x = Rs. 14,000. Then,

z = (14,000 - 10,000)/2,000 = 2

This means that the salary of Rs. 14,000 is 2 standard deviations above the mean.

Step 2: Find the area under the curve
We need to find the area under the curve to the right of z = 2, which represents the proportion of workers receiving salary more than Rs. 14,000. We can use a standard normal distribution table or a calculator to find this area.

Using a standard normal distribution table, we find that the area to the right of z = 2 is 0.0228.

Step 3: Calculate the total number of workers
Let N be the total number of workers in the factory. Then, the number of workers receiving salary more than Rs. 14,000 is given by:

50 = N * 0.0228

Solving for N, we get:

N = 50/0.0228 = 2192.98

Rounding off to the nearest integer, we get:

N = 2193

Therefore, the total number of workers in the factory is 2,193 (option A).

Uniform Distribution

The distribution of points obtained in a single throw of an unbiased die is an example of uniform distribution. This is because each of the six faces of the die has an equal probability of appearing, i.e., 1/6. Therefore, the probability of obtaining any particular number between 1 and 6 is the same, and the distribution of points is uniform.

Definition of Uniform Distribution

Uniform distribution is a probability distribution where every possible outcome has an equal chance of occurring. In other words, the probability of any event is the same, and the distribution is flat or uniform. A uniform distribution is often used to model situations where every outcome is equally likely.

Characteristics of Uniform Distribution

Some of the characteristics of the uniform distribution are:

1. The probability density function (PDF) is constant over the range of possible values.

2. The mean and variance of the distribution can be easily calculated.

3. The distribution is symmetric.

4. The distribution is defined by its minimum and maximum values.

Conclusion

In conclusion, the distribution of points obtained in a single throw of an unbiased die follows a uniform distribution. This is because each of the six faces of the die has an equal probability of appearing, and the distribution is flat or uniform. The characteristics of the uniform distribution include a constant PDF, easy calculation of mean and variance, symmetry, and definition by minimum and maximum values.

Solution:
Given, probability of recovering = 0.75
Number of patients = 48

1. Expected Value:
The expected value of the number of recoveries can be calculated as follows:
Expected Value = Number of patients × Probability of recovering
Expected Value = 48 × 0.75
Expected Value = 36

2. Variance:
The variance of the number of recoveries can be calculated as follows:
Variance = Number of patients × Probability of success × Probability of failure
Variance = 48 × 0.75 × 0.25
Variance = 9

3. Standard Deviation:
The standard deviation can be calculated by taking the square root of the variance:
Standard Deviation = √(Variance)
Standard Deviation = √(9)
Standard Deviation = 3

Therefore, the correct answer is option 'D' (3).

Explanation:

Binomial Distribution: A binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters or assumptions. The binomial distribution formula is:

P(X=k) = C(n,k) * p^k * (1-p)^(n-k)

Where,

P(X=k) is the probability of getting ‘k’ successes in ‘n’ trials,
C(n,k) is the number of possible combinations of ‘n’ trials where ‘k’ successes are observed,
p is the probability of success in any given trial,
(1-p) is the probability of failure in any given trial,
n is the total number of trials.

Variance of Binomial Distribution: The variance of a binomial distribution is given by:

Var(X) = np(1-p)

Where,

n is the number of trials,
p is the probability of success in any given trial.

Maximum Variance of Binomial Distribution: To find the maximum value of the variance of a binomial distribution, we take the first derivative of the variance with respect to p and equate it to zero. This gives us the maximum value of the variance as:

p = 1/2

Substituting this value of p in the variance formula, we get:

Var(X) = n/4

Therefore, the maximum value of the variance of a binomial distribution with parameters n and p is n/4.

Answer: The correct answer is option 'B'.

Solution:

Probability of a missile hitting a target
Given that the probability of a missile hitting a target is 1/8.

Therefore, the probability of missing the target is (1-1/8) = 7/8.

Probability of at least 2 missiles hitting the target out of 10 missiles fired
We need to find the probability of at least 2 missiles hitting the target out of 10 missiles fired.

Let X be the number of missiles that hit the target out of 10 missiles fired.

Therefore, X follows a binomial distribution with n = 10 and p = 1/8.

Using the formula for the probability mass function of binomial distribution, we have:

P(X ≥ 2) = 1 - P(X < />

= 1 - P(X = 0) - P(X = 1)

= 1 - (10C0) (1/8)^0 (7/8)^10 - (10C1) (1/8)^1 (7/8)^9

= 1 - (1) (1) (0.4783) - (10) (0.125) (0.5369)

= 0.3611

Therefore, the probability that out of 10 missiles fired, at least 2 will hit the target is 0.3611.

Hence, option (d) is the correct answer.

Solution:

Given, an unbiased die is tossed 500 times.

Let X be the number of sixes in 500 tosses.

We know that the probability of getting a six in a single toss of an unbiased die is 1/6.

The random variable X follows a binomial distribution with parameters n = 500 and p = 1/6.

The mean of a binomial distribution is given by np.

Therefore, the mean of X is given by np = 500*(1/6) = 500/6.

Hence, the correct option is (B) 500/6.

Mean μ and standard deviation σ, then we can write:

X ~ N(μ, σ)

This notation means that X follows a normal distribution with mean μ and standard deviation σ. The symbol "~" is read as "is distributed as" or "follows".

For example, if the height of a group of people follows a normal distribution with a mean of 170 cm and a standard deviation of 5 cm, we can write:

height ~ N(170, 5)

This means that the height of a person from this group is distributed as a normal distribution with mean 170 cm and standard deviation 5 cm.

Method of Moments for Fitting a Binomial Distribution

The method usually applied for fitting a binomial distribution is known as the method of moments. This method involves equating the sample moments of the distribution to their corresponding population moments.

Steps Involved in Applying the Method of Moments for Fitting a Binomial Distribution:

1. Determine the Sample Moments:
The first step is to determine the sample moments of the distribution. For a binomial distribution, the first two sample moments are the sample mean and the sample variance. These can be calculated from the data.

2. Determine the Population Moments:
The next step is to determine the corresponding population moments. For a binomial distribution, the first two population moments are the population mean and the population variance.

3. Equate the Sample and Population Moments:
The final step is to equate the sample and population moments. This will result in a set of equations that can be solved for the parameters of the binomial distribution.

When applying the method of moments to a binomial distribution, the parameter p is estimated by equating the sample mean to the population mean, and the parameter n is estimated by equating the sample variance to the population variance.

Limitations of the Method of Moments:

1. The method of moments may not always provide accurate estimates of the parameters of a distribution, especially for small sample sizes.

2. The method assumes that the distribution is well-behaved and has finite moments.

3. The method does not take into account the shape of the distribution, and may not be appropriate for distributions with multiple modes or skewness.

Conclusion:

In conclusion, the method of moments is a commonly used method for fitting a binomial distribution. It involves equating the sample moments to their corresponding population moments, and solving for the parameters of the distribution. However, the method has its limitations and should be used with caution, especially for small sample sizes.

Binomial Distribution and Infinite n

Binomial Distribution is a probability distribution that describes the number of successes in a fixed number of independent trials. The probability of success in each trial is denoted by p, and the probability of failure is denoted by q = 1-p.

As the number of trials, n, becomes infinitely large, there are some observations:

- The probability of occurrence of the event, p, approaches 0.
- The probability of non-occurrence of the event, q, approaches 1.

Explanation

When n is very large, the probability of occurrence of the event in each trial becomes very small. As a result, the total number of successes in the n trials will also be small. This means that the binomial distribution will be skewed towards 0.

As n approaches infinity, the binomial distribution approaches a normal distribution. In a normal distribution, the mean is equal to the probability of occurrence of the event, p. As the probability of occurrence of the event approaches 0, the mean also approaches 0.

Similarly, in a normal distribution, the standard deviation is equal to the square root of pq. As q approaches 1, the standard deviation also approaches 0. This means that the distribution becomes more and more concentrated around the mean.

Conclusion

In summary, as the number of trials, n, becomes infinitely large, the probability of occurrence of the event, p, approaches 0, and the probability of non-occurrence of the event, q, approaches 1. This makes the binomial distribution skewed towards 0 and more and more concentrated around the mean.

Binomial Distribution:

Binomial Distribution is a discrete probability distribution that deals with the number of successes in a fixed number of independent trials [1]. It is usually denoted as B(n, p), where n is the number of trials and p is the probability of success in each trial.

Mean and Standard Deviation of Binomial Distribution:

Mean and Standard Deviation of Binomial Distribution can be calculated using the following formulas [2]:

Mean (μ) = np

Standard Deviation (σ) = √(npq)

where q = 1 - p

Solution:

Given, Mean (μ) = 20, Standard Deviation (σ) = 4

To find q, we need to use the formula of Standard Deviation of Binomial Distribution:

σ = √(npq)

Squaring both sides, we get:

σ^2 = npq

Substituting the given values, we get:

4^2 = 20q(1 - q)

16 = 20q - 20q^2

20q^2 - 20q + 16 = 0

Dividing both sides by 4, we get:

5q^2 - 5q + 4 = 0

Using the quadratic formula, we get:

q = [5 ± √(5^2 - 4*5*4)]/(2*5)

q = [5 ± √(25 - 80)]/10

q = [5 ± √(-55)]/10

As q has to be between 0 and 1, and we can't take the square root of a negative number, we discard the negative solution.

q = [5 + √(-55)]/10

q = [5 + i√55]/10

where i is the imaginary unit (√-1)

Therefore, q is equal to 4/5 (Option D).

Conclusion:

Binomial Distribution is a useful tool in probability and statistics, which helps in calculating the number of successes in a fixed number of independent trials. The mean and standard deviation of binomial distribution can be calculated using the appropriate formulas. In this question, we used the standard deviation formula to find q, and after discarding the negative solution, we arrived at the correct answer of q = 4/5.

Given: Points of inflexion of a normal curve are 40 and 60 respectively.

To find: Mean deviation.

Solution:

We know that the points of inflexion of a normal curve are symmetric about the mean.

Let the mean of the normal curve be 'M'.

So, the distance of the points of inflexion from the mean would be:

(60 - M) = (M - 40)

Solving this equation, we get:

M = 50

So, the mean of the normal curve is 50.

Now, we need to find the mean deviation.

Mean deviation is given by the formula:

Mean deviation = ∫|x - M| f(x) dx / ∫f(x) dx

Where, f(x) is the probability density function of the normal curve.

For a standard normal curve, the mean deviation is 1.

But in this case, we need to adjust the formula for the mean deviation based on the given mean 'M'.

So, the adjusted formula for mean deviation is:

Mean deviation = ∫|x - 50| f(x) dx / ∫f(x) dx

Now, we can use the properties of the normal curve to simplify this formula.

We know that the normal curve is symmetric about the mean.

So, we can split the integral into two parts:

∫|x - 50| f(x) dx = ∫(x - 50) f(x) dx for x > 50

+ ∫(50 - x) f(x) dx for x < />

Since the normal curve is a continuous probability distribution, we can assume that the probability density function is continuous and differentiable.

So, we can use the fact that the derivative of the normal curve is proportional to the distance from the mean.

So, we have:

f'(x) = k (x - 50) f(x) for x > 50

f'(x) = -k (x - 50) f(x) for x < />

Where, k is a constant of proportionality.

Integrating both sides, we get:

ln(f(x)) = k/2 (x - 50)^2 + C for x > 50

ln(f(x)) = -k/2 (x - 50)^2 + C for x < />

Where, C is a constant of integration.

Taking the exponential of both sides, we get:

f(x) = A exp(k/2 (x - 50)^2) for x > 50

f(x) = A exp(-k/2 (x - 50)^2) for x < />

Where, A is a constant of proportionality.

Now, we can use these formulas to evaluate the integrals.

∫(x - 50) f(x) dx for x > 50

= ∫(x - 50) A exp(k/2 (x - 50)^2) dx for x > 50

= A/k exp(k/2 (x - 50)^2) + C1 for x > 50

Where, C1 is a constant of integration.

∫(50 - x) f(x) dx for x < />

= ∫(50 - x) A exp(-k/2 (x - 50)^2) dx for x < />

Meaning of Binomial Distribution

Before we move ahead to solve the problem, let's first understand the concept of binomial distribution.

A binomial distribution is a probability distribution that describes the number of successes in a fixed number of independent trials, where the probability of success remains the same in all the trials. The distribution can be represented by two parameters - n and p, where n is the number of trials and p is the probability of success in each trial.

The mean of a binomial distribution is given by the formula:

μ = n × p

The standard deviation of a binomial distribution is given by the formula:

σ = √(n × p × (1 - p))

Solving the Problem

Now, let's solve the given problem step by step.

Given,

Mean (μ) = 3

Standard deviation (σ) = 1.5

We know that,

μ = n × p

σ = √(n × p × (1 - p))

Let's substitute the given values in these formulas and solve for n and p.

From the first equation, we get:

p = μ / n

Substituting the value of μ and p, we get:

3 = n × (μ / n)

3 = μ

So, n = 3 / μ = 3 / 3 = 1

Now, let's substitute the values of n and μ in the second equation:

σ = √(n × p × (1 - p))

1.5 = √(1 × p × (1 - p))

Squaring both sides, we get:

2.25 = p × (1 - p)

2.25 = p - p^2

Rearranging, we get:

p^2 - p + 2.25 = 0

Solving this quadratic equation, we get:

p = 1.5, 0.5

Since the probability of success cannot be greater than 1, we reject the value p = 1.5.

Therefore, p = 0.5

Now, we know that n = 1 and p = 0.5.

The number of trials in a binomial distribution is given by n. Therefore, the answer is:

No. of trials (n) = 3 / μ = 3 / 3 = 1 / p = 1 / 0.5 = 2

But, the question asks for the number of trials in a binomial distribution having mean and SD as 3 and 1.5 respectively. Since we have found the values of n and p, we can check if they satisfy the given conditions.

Mean (μ) = n × p = 1 × 0.5 = 0.5 ≠ 3

Standard deviation (σ) = √(n × p × (1 - p)) = √(1 × 0.5 × 0.5) = 0.5 ≠ 1.5

Therefore, the values of n and p that we have found do not satisfy the given conditions.

We need to go back and check our calculations.

From the equation p^2 - p + 2.25 = 0, we get two values of p:

p = 1.5, -0.5

We reject the negative value of p.

Therefore, p =

Binomial Distribution. A binomial random variable is the number of successes x in n repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a binomial distribution. Suppose we flip a coin two times and count the number of heads (successes).

The Poisson distribution has a single parameter, λ . For a Poisson distribution modeling a binomial phenomenon, λ can be taken as an approximation of np.In such a case, a Poisson distribution with the appropriate parameter λ will approximate the distribution of events over time or the number of events in an interval.is unimodal.and multi-modal..

Explanation:
A standard normal distribution refers to a normal distribution with a mean of 0 and a standard deviation of 1. The points of inflexion on a normal distribution curve occur where the curve changes from being concave upwards to concave downwards, or vice versa. These points are located at one standard deviation away from the mean on either side of the curve.

Therefore, the correct answer is option 'C', where the points of inflexion are located at 1 standard deviation away from the mean on either side of the curve, which is represented by the values of x = 1 and x = -1.

To summarize:
- A standard normal distribution has a mean of 0 and a standard deviation of 1.
- Points of inflexion on a normal distribution curve occur where the curve changes from being concave upwards to concave downwards, or vice versa.
- The points of inflexion on a standard normal distribution curve are located at 1 standard deviation away from the mean on either side of the curve, which is represented by the values of x = 1 and x = -1.

There are 4 values less than equal to 15.. i.e 8,9,11,15.. so probability is: 4/6=2/3.

Explanation:

Given, x follows a uniform distribution and takes values 8, 9, 11, 15, 18, 20.

As x follows a uniform distribution, each value has an equal probability of occurring.

Hence, P(x = 8) = P(x = 9) = P(x = 11) = P(x = 15) = P(x = 18) = P(x = 20) = 1/6.

As 12 is not one of the values that x can take, P(x = 12) = 0.

Therefore, option B is the correct answer.


Answer: B) 0

Frequency Function of an Unbiased Die

An unbiased die is one in which each face has an equal probability of landing face up. The frequency function of an unbiased die is a function that gives the probability of each possible outcome, which is the number that appears on the top face of the die.

Options for the Frequency Function

a) f(x) = 1/4
b) f(x) = 1/5
c) f(x) = 1/6
d) none

Explanation of Correct Answer

The correct answer is option C, f(x) = 1/6. This is because there are six possible outcomes when rolling an unbiased die, which are the numbers 1, 2, 3, 4, 5, and 6. Since each outcome has an equal probability of occurring, the probability of each outcome is 1/6.

Therefore, the frequency function of an unbiased die is f(x) = 1/6, which means that the probability of rolling any specific number on the die is 1/6. This is the most likely outcome for rolling an unbiased die, and it is the correct answer to this question.

Probability distribution can be classified into two categories, discrete and continuous. However, probability distribution can also be infinite, which means there are infinite possible outcomes. Therefore, the correct answer is option D, that is, probability distribution can be both discrete and continuous.

Discrete Probability Distribution:
Discrete probability distribution deals with the probability of occurrence of discrete random variables. Discrete random variables are the variables that can only take on a finite or countable number of values. For example, the number of heads that can be obtained when flipping a coin is a discrete random variable. The most common examples of discrete probability distributions are the binomial distribution, Poisson distribution, and geometric distribution.

Continuous Probability Distribution:
Continuous probability distribution deals with the probability of occurrence of continuous random variables. Continuous random variables are the variables that can take on an infinite number of values within a given range. For example, the time taken by a car to travel from one point to another is a continuous random variable. The most common examples of continuous probability distributions are the normal distribution, exponential distribution, and uniform distribution.

Infinite Probability Distribution:
Infinite probability distribution deals with the probability of occurrence of infinite random variables. Infinite random variables are the variables that can take on an infinite number of values without any limit. For example, the height of people in a given population is an infinite random variable. The most common examples of infinite probability distributions are Cauchy distribution, Pareto distribution, and Levy distribution.

Conclusion:
Probability distribution can be classified into discrete, continuous, and infinite probability distributions. Discrete probability distribution deals with the probability of occurrence of discrete random variables, continuous probability distribution deals with the probability of occurrence of continuous random variables, and infinite probability distribution deals with the probability of occurrence of infinite random variables. Therefore, the correct answer is option D, that is, probability distribution can be both discrete and continuous.

= 2)?

Let the number of trials be n and the probability of success in each trial be p.
Then, the mean of the binomial distribution is given by np = 5.
We are also given that 10P(X=0) = P(X=1).
Using the formula for the probability mass function of the binomial distribution, we have:

10P(X=0) = P(X=1)
10(1-p)^n = np(1-p)^(n-1)

Dividing both sides by (1-p)^(n-1), we get:

10(1-p) = np

Substituting np = 5, we have:

10(1-p) = 5
1-p = 0.5
p = 0.5

Now, using the formula for the probability mass function of the binomial distribution, we have:

P(X=2) = (n choose 2) p^2 (1-p)^(n-2)

Since we do not know the value of n, we cannot find the exact value of P(X=2).
However, we can use the fact that np = 5 to estimate n.
We have:

np = 5
n = 10

Substituting n=10 and p=0.5, we have:

P(X=2) = (10 choose 2) (0.5)^2 (0.5)^(10-2) ≈ 0.246

Therefore, the value of P(X=2) is approximately 0.246.

Poisson distribution is a statistical concept that describes the probability of a given number of events occurring in a fixed interval of time or space. It is used to model the number of times that an event occurs in a given time frame, such as the number of defects in a product or the number of customers who arrive at a store.

Explanation:

In this given problem, we are interested in the number of radioactive atoms that decay in a given interval of time. The decay of radioactive atoms is a random process, and the rate of decay is proportional to the number of atoms present. Therefore, the number of decays that occur in a fixed interval of time follows a Poisson distribution.

The Poisson distribution has the following properties:

- It is a discrete probability distribution that ranges from 0 to infinity.
- The mean and variance of the distribution are equal and are denoted by λ.
- The probability of observing k events in a fixed interval of time is given by the formula:

P(k) = (e^(-λ) * λ^k) / k!

where e is the mathematical constant approximately equal to 2.71828.

In the given problem, the number of decays that occur in a fixed interval of time can be modeled by a Poisson distribution with a mean of λ. The probability of observing k decays in the interval is given by the Poisson distribution formula.

Therefore, the correct answer is option C, Poisson distribution.

Binomial Distribution

Binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials. It has two parameters - n (number of trials) and p (probability of success in each trial).

Mean and Variance of Binomial Distribution

The mean (μ) and variance (σ²) of the binomial distribution are given by:

μ = np

σ² = np(1-p)

Given that the mean of x is 2 and the variance is 1.6, we can write:

μ = np = 2

σ² = np(1-p) = 1.6

Solving these equations, we get:

p = μ/n = 2/n

σ² = np(1-p) = 1.6

Substituting the value of p in the second equation, we get:

1.6 = 2/n * (1 - 2/n)

Simplifying this equation, we get:

n² - 5n + 8 = 0

Solving this quadratic equation, we get:

n = 4 or n = 1

Since n is the number of trials, it cannot be 1. Therefore, n = 4.

Substituting the value of n in the expression for p, we get:

p = 2/4 = 1/2

Therefore, the correct option is A) 1/5.

Explanation:

Option A is the correct answer because it corresponds to the value of p = 1/2 - which we obtained from the mean and variance of the binomial distribution. The other options are incorrect because they do not correspond to the calculated value of p.

Given information
- An unbiased die is tossed 500 times.
- We need to find the standard deviation of the number of sixes in these 500 tosses.

Approach
- We can find the standard deviation using the formula for a binomial distribution.
- In a binomial distribution, the probability of success (getting a six) is p = 1/6, and the probability of failure (not getting a six) is q = 5/6.
- The mean of a binomial distribution is np, where n is the number of trials and p is the probability of success.
- The variance of a binomial distribution is npq.
- The standard deviation is the square root of the variance.

Calculations
- n = 500 (number of trials)
- p = 1/6 (probability of success)
- q = 5/6 (probability of failure)
- Mean = np = 500 * 1/6 = 83.33
- Variance = npq = 500 * 1/6 * 5/6 = 69.44
- Standard deviation = square root of variance = sqrt(69.44) = 8.33

Answer
- The standard deviation of the number of sixes in 500 tosses of an unbiased die is 8.33.
- Option A (50/6) is the closest answer to the calculated standard deviation (8.33).

Distribution of (X Y) when X and Y are independent random variables

When two random variables X and Y are independent, the joint probability distribution of (X, Y) is the product of their marginal probability distributions.

Thus, if X ~ 2m and Y ~ 2n, then the joint probability distribution of (X, Y) is:

f(X, Y) = f(X) * f(Y)

where f(X) and f(Y) are the probability density functions of X and Y, respectively.

Finding the distribution of (X Y)

To find the distribution of (X Y), we need to find the probability density function of Z = X Y.

Let F(Z) be the cumulative distribution function of Z. Then:

F(Z) = P(Z ≤ z) = P(X Y ≤ z)

Since X and Y are independent, we can express the joint probability distribution of (X, Y) as:

f(X, Y) = f(X) * f(Y)

Thus, we can rewrite the probability density function of Z as:

f(Z) = ∫∫ f(X, Y) * δ(X Y - z) dX dY

where δ is the Dirac delta function.

Solving the integral, we get:

f(Z) = ∫∫ f(X) * f(Y) * δ(X Y - z) dX dY

= ∫∞-∞ f(X) * f(z/X) / |X| dX

where |X| denotes the absolute value of X.

This is the probability density function of Z, which is a chi-square distribution with 2 degrees of freedom.

Therefore, the distribution of (X Y) is a chi-square distribution with 2 degrees of freedom when X and Y are independent random variables with X ~ 2m and Y ~ 2n.

Binomial Distribution and Sum of Powers of p and q

Binomial distribution is a type of probability distribution that describes the probability of a certain number of successes in a fixed number of independent trials. The distribution is characterized by two parameters - the probability of success (p) and the probability of failure (q), where q = 1 - p.

The sum of powers of p and q is always equal to n, where n is the number of independent trials. This property holds true for all values of the number of successes.

Proof:

Let X be a random variable that follows a binomial distribution with parameters n and p. Then, the probability mass function of X is given by:

P(X = k) = nCk p^k q^(n-k)

where nCk is the binomial coefficient.

The sum of powers of p and q can be expressed as:

p^n + q^n = (p+q)^n

Using the binomial theorem, we can expand (p+q)^n as:

(p+q)^n = ∑_(k=0)^n nCk p^k q^(n-k)

Comparing this with the probability mass function of X, we can see that the sum of powers of p and q is equal to the sum of the probabilities of all possible outcomes of X. Since the probabilities of all possible outcomes of X add up to 1, we have:

p^n + q^n = (p+q)^n = 1^n = 1

Therefore, the sum of powers of p and q is always equal to n, regardless of the number of successes.

Conclusion:

Hence, we can conclude that the statement "For n independent trials in Binomial distribution the sum of the powers of p and q is always n , whatever be the no. of success" is true.

Given:

- X is a binomial variable
- 2 P(X = 2) = P(X = 3)
- Mean of X = 10/3

To find:

- Probability that X assumes at most the value 2

Solution:

1. Using the mean of X:

- Mean of X = np, where n is the number of trials and p is the probability of success in each trial
- np = 10/3
- p = (10/3)/n

2. Using the given relationship between P(X = 2) and P(X = 3):

- 2 P(X = 2) = P(X = 3)
- 2 (n C 2) p^2 (1-p)^(n-2) = (n C 3) p^3 (1-p)^(n-3)
- Simplifying and solving for p: p = 2/3

3. Using p = (10/3)/n:

- (10/3)/n = 2/3
- n = 5

4. Using the binomial distribution formula:

- P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
- P(X = k) = (n C k) p^k (1-p)^(n-k)
- P(X = 0) = (5 C 0) (2/3)^0 (1/3)^5 = 1/243
- P(X = 1) = (5 C 1) (2/3)^1 (1/3)^4 = 10/243
- P(X = 2) = (5 C 2) (2/3)^2 (1/3)^3 = 40/243
- P(X ≤ 2) = 1/243 + 10/243 + 40/243 = 17/81

Therefore, the probability that X assumes at most the value 2 is 17/81, which is option (B).

Poisson distribution ka formula lagao then x ki value daalo fir solve kro .... aajayega..... I'm not able to send photo,,, I would've sent it

Given data:
Average salary (μ) = Rs.10,000
Standard deviation (σ) = Rs.2,000

Let X be the salary of a worker in the factory. Then,

X ~ N(10000, 2000^2)

To find the number of workers receiving a salary more than Rs.14,000, we need to find the probability of X > 14000.

P(X > 14000) = P(Z > (14000-10000)/2000) [where Z is the standard normal variable]

= P(Z > 2)

= 0.0228 (from standard normal table)

Now, we are given that 50 workers receive a salary more than Rs.14,000. Let the total number of workers in the factory be N.

Then, the number of workers receiving a salary less than or equal to Rs.14,000 is N - 50.

Using the normal distribution, we can write:

P(X ≤ 14000) = P(Z ≤ (14000-10000)/2000)

= P(Z ≤ 2)

= 0.9772

We know that the area under the normal curve is equal to 1. Hence,

P(X ≤ 14000) + P(X > 14000) = 1

0.9772 + 0.0228 = 1

So, the proportion of workers receiving a salary less than or equal to Rs.14,000 is 0.9772.

Therefore, (N - 50)/N = 0.9772

Solving for N, we get N = 2193.

Hence, the total number of workers in the factory is 2193.

Therefore, the correct option is (a) 2,193.