MCQs' Theoretical Distributions - Quantitative Aptitude for CA Foundation

Q1. When p = 0.5, the distribution is
(a) Asymmetrical
(b) Symmetrical
(c) Both of above
(d) None of above
Ans: (b)

Sol:

We know,

Binomial distribution is symmetric when p = 0.5

Q2. If mean and standard deviation of a binomial distribution is 10 and 2 respectively, q will be __
(a) 1
(b) 0.8
(c) 0.6
(d) 0.4
Ans: (d)

Sol:

According to the question, we have

Mean = 10 & S.D = 2
i.e., m = np = 10, S.D. =  √npq = 2

Dividing both:

npq/np= 4/10
q = 0.4

Q3. Find the variance of binomial distribution with n = 10, p = 0.3
(a) 2.1
(b) 3
(c) 7
(d) None of these
Ans: (a)

Sol:
Given, n = 10 & p = 0.3

Thus, q = 1 – p = 1 – 0.3 = 0.7

Therefore, Variance = npq = 10 × 0.3 × 0.7 = 2.1

Q4. What is the probability of making 3 corrected guesses in 5 True–False answer type questions
(a) 0.3125
(b) 0.4156
(c) 1.3888
(d) 0.5235
Ans: (a)

Sol:

According to the question,

Number of trials; n = 500

Specific outcome (making 3 correct guesses): x = 3

Probability of success: P= 1/2

Probability of failure:  q = 1

Binomial distribution is given by the formula,

10/32 = 5/16 = 0.3125

Q5. What will be the mode of the Binomial Distribution in which mean is 20 and standard deviation is √10 ?
(a) 21
(b) 20.5
(c) 20
(d) 41
Ans: (c)

Sol:
Given, np = 20)  ...(i)
⇒ √npq  =  √10 ...(ii)

Put this value in eq. (i), we get

20q = 10

⇒ q = 1/2

Thus, l = 1/2

Also, np = 20

⇒ n (1/2) = 20

⇒ n = 40

Now, Mode = (n + 1) l = 41 × 1/2 = 20.5

Since, it is a non-integer, thus

Mode = 20

Q6: The overall percentage of failure in a certain examination is 0.30. What is the probability that out of a group of 6 candidates at least 4 passed the examination? 
(a) 0.74
(b) 0.71
(c) 0.59 
(d) 0.67
Ans: (a)

Sol:
According to question, we have

q = 0.3, p = 1 – 0.3 = 0.7 and n = 6

Therefore, the probability that out of a group of

6 candidates at least 4 passed the examination is given by;

P(x ≥ 4) = ⁶C₄(0.7)⁴ (0.3)² + ⁶C₅(0.7)⁵ (0.3)¹ + ⁶C₆₍0.7)⁶ (0.3)⁰

⇒ P(x ≥ 4) = 0.324135 + 0.30256  + 0.117649

⇒ P(x ≥ 4) = 0.74 (Approx)

Q7: Examine the validity of the following : Mean and standard deviation of a binomial distribution are 10 and 4 respective:
(a) Valid
(b) Not valid
(c) Both (a) and (b)
(d) Neither (a) nor (b)
Ans: (b)

Sol:
According to question, we have

Mean = 10

S.D = 4

Variance = 16

Since, mean is always greater than variance.

But 10 < 16

Hence, the given data is not valid.

Q8: Number of misprints per page of a thick book follows;
(a) Normal distribution
(b) Poisson distribution
(c) Binomial distribution
(d) Standard normal distribution
Ans: (b)

Sol:

Since, the number of trials (n) are very large and tends to be infinite however success (p) is very small.

Therefore, number of misprints per page   of a thick book follows Poisson distribution.

Q9: In _____ distribution, mean = variance.
(a) Normal
(b) Binomial
(c) Poisson
(d) None of these
Ans: (c)

Sol:
For Poisson distribution, mean = variance.

Q10: Shape of Normal Distribution Curve:
(a) Depends on its parameters
(b) Does not depend on its parameters
(c) Either (a) or (b)
(d) Neither (a) nor (b)
Ans: (a)

Sol:

We know,

Shape of Normal Distribution curve depends on its parameters.

Q11: If 3 percent of ceramic cup manufactured by a company are known to be defective. What is the probability that a sample of 100 cups are taken from the production process, of that company would contain exactly one defective cup?
(a) 0.03
(b) 0.15
(c) 0.09
(d) 0.30
Ans: (b)

Sol:

Given: p = 3/100 = 0.03 and n = 100

Thus, np = n × p = 3/100 × 100 = 3

Now, the required probability is given by:

P(X = 1) = e⁻³ × 3¹/1!

⇒ P(X = 1) = e⁻³ × 3

⇒ P(X = 1) = 0.149 ≈ 0.15

Q12: If 5% of the families in large population city do not used gas as a fuel, what will be the probability of selecting 10 families in a random sample of 100 families who do not use gas as a fuel?
(Given that e^-5 = 0.0067)
(a) 0.038
(b) Zero
(c) 0.018
(d) 0.048
Ans: (c)

Sol:
Given:

• p = 5/100 = 0.05
• n = 100
• So, np = 5

Using Poisson distribution, we get

P(x = 10) = e⁻⁵ · 5¹⁰/10!

e⁻⁵ = 0.0067

P(x = 10) = 0.0067 · 9765625/3628800 ≈ 0.01803

Q13: The number of accidents in a year attributed to taxi drivers in a locality follows Poisson distribution with average 2. Out of 500 taxi drivers of that area, what is the number of drivers with at least 3 accidents in a year?
(Given that e = 2.718)
(a) 162
(b) 180
(c) 201
(d) 190
Ans: (a)

Sol:
Given,
Total no. of taxi drivers (N) = 500
Mean (m) = 2
The probability of at least 3 accidents is given by
P(X ≥ 3) = 1 – P(X < 3)
We know that, P (X = x) = e^-mm^x/x!
⇒ P(X ≥ 3) = 1 – P(X < 3)
⇒ P(X ≥ 3) = 1 – [P(0) + P(1) + P(2)]
⇒ P(X ≥ 3) = 1 –

⇒ P(X ≥ 3) = 1 –

⇒ P(X ≥ 3) = 1 –
⇒ P(X ≥ 3) = 1 – 0.06767
⇒ P(X ≥ 3) = 0.3233

Now, the number of drivers with at least 3 accidents in a year,
= N × P(X ≥ 3)
= 500 × 0.3233
= 161.65 ≈ 162
Therefore, the number of driver with at least 3 accidents in a year are 162.

Q14: The probability than a man aged 45 years will die within a year is 0.012. What is the probability that of 10 men, at least 9 will reach their 46th birthday? [Given; e^–0.12 = 0.88692]
(a) 0.0935 
(b) 0.9934
(c) 0.9335 
(d) 0.9555
Ans: (c)

Sol:
According to the question, we have

p = 0.012

n = 10

m = np

 = 10 × 0.012 = 0.12

Now, the probability that at least 9 men survive

i.e.,

9 reached/survive or 10 reached/survive

or

1 died or 0 died

= P(X = 0) + P(X = 1)

= 0.88692[1 + 0.12]
= 0.9934 (Approx)

Q15: An example of bi‑parametric continuous probability distribution.
(a) Binomial
(b) Poisson
(c) Normal
(d) (a) and (b)
Ans: (a)

Sol:
A bi-parametric continuous distributionmeans it has two parameters and is continuous

• Binomial → discrete
• Poisson → discrete
• Normal
• [μ, σ²]

Q16: In a normal distribution, skewness is __.
(a) 0
(b) > 3
(c) < 3
(d) < 1
Ans: (a)

Sol:
We know,

Normal distribution is perfectly symmetric, so skewness = 0.

Q17: The total area of the normal curve is the
(a) 50 percent
(b) one
(c) 0.50
(d) any value between 0 and 1
Ans: (b)

Sol:
The total probability under a normal curve equals 1.

Q18: If X and Y are two independent normal variables with means 10 and 12, and standard deviations (S.D.) 3 and 4 respectively, then (X + Y) is normally distributed with:
(a) Mean = 22 and S.D. = 7
(b) Mean = 22 and S.D. = 25
(c) Mean = 22 and S.D. = 5
(d) Mean = 22 and S.D. = 49
Ans: (c)

Sol:
Means of X + Y:

• μ = 10 + 12 = 22

Standard deviation of X + Y (since independent variables):

Q19: If the points of inflexion of a normal curve are 40 and 60 respectively, then its mean is
(a) 40
(b) 45
(c) 50
(d) 60
Ans: (c)

Sol:
Given: Points of inflexion of normal curve are 40 and 60.
Let σ be the standard deviation and μ be the mean, then According to the question,
μ – σ = 40 …(i)
μ + σ = 60 …(ii)
Adding (i) and (ii),

⇒ μ = 100/2
⇒ μ = 50
Therefore, the required mean is 50.

Q20: X follow normal distribution with mean as 50 and variance as 100. What is P(X ≥ 60)?
[Given φ(1) = 0.8413]
(a) 0.20
(b) 0.40
(c) 0.16
(d) 0.30
Ans: (c)

Sol:
According to question, we have
μ = 50 and σ² = 100
⇒ σ = 10
Z = X − μ/σ
Z = x − 50/10

P(x ≥ 60) = P  
= P(Z ≥ 1)
= 1 − 0.8413
= 0.1587