Statistics- 1 - Free MCQ Practice Test with solutions, JEE Maths

Arranging the given data in ascending order,
we have 34, 38, 42, 44, 46, 48, 54, 55, 63, 70
Here, Median M = (46+48)/2
=47
(∵ n = 10, median is the mean of 5th and 6th items)
∴ Mean deviation = ∑|xi−M|/n
=∑|xi−47|/10
= (13+9+5+3+1+1+7+8+16+23)/10
=8.6

μ = (22 + 26 + 28 + 20 + 24 + 30)/6
= 150/6 
= 25
x(i) = (xi - μ)²
x(22) = (22-25)² = 9
x(26) = (26-25)² = 1
x(28) = (28-25)² = 9
x(20) = (20-25)² = 25
x(24) = (24-25)² = 1
x(30) = (30-25)² = 25
​
(xi - μ)²  = 70
Standard deviation : [(xi - μ)²]/N
= (70/6)^½
= 3.42

ρ = (b(xy) * b(yx))
But sign of ρρ is same as sign of b(xy), b(yx)
Therefore, ρ = 2/7

Given lines are x+4y=3   (i) 
& 3x+y=15        (ii)
We first check which of the line (i) & (ii) is line of regression y
on x and x on y. Let line x+4y=3 be the line of regression x on y, then
other be the line of regression y on x.
∴ From (i) i.e. line of regression x on y we have x=−4y+3
∴ b(xy)=regression coefficient x on y=−4 and 3x+y=15
∴ y=−3x+15
∴ b(yx)=regression coefficient y on x=-3
Now count r²=b(yx)b(xy)=(−4)(−3)=12
⇒ ∣r∣=2(3)^1/2
​which is not possible as 0≤r²≤1. So our assumption is wrong
and line x+4y=3 is line of regression y on x & 3x+y=15 is line of
regression x on y.
∴ 3x+y=15
⇒ x=(15−y)/3
⇒ x=(15−3)/3   (By putting y=3)
​= 4

Let the other nos. be a and b
then (x+y+1+2+6)/5 = 4.4
x + y = 13 ---------------------(1)
Variance = 8.24


41.2 = 19.88 + (x2 + 19.36 – 8.8x) + (y2 + 19.36 – 8.8y) 
21.32 = x2 + y2 + 38.72 – 8.8(x + y) 
x2 + y2 + 38.72 – 8.8(13) – 21.32 = 0 
(using equation (1)) 
x2 + y2 – 97 = 0 …(2) 
Squaring equation (1) both the sides, 
we get (x + y)2 = (13)^2 
x2 + y2 + 2xy = 169 
97 + 2xy = 169 
(using equation (2)) 
xy = 36 or x = 36/y (1)
⇒ 36/y + y = 13 
y2 + 36 = 13y 
y2 – 13y + 36 = 0 
(y – 4)(y – 9) = 0 
Either (y – 4) = 0 or (y – 9) = 0 
⇒ y = 4 or y = 9 
For y = 4 x = 36/y 
= 36/4 = 12 
For y = 9 
x = 36/9 
x = 4 
Thus, remaining two observations are 4 and 9.

The relationship between the mean, quartile and the standard deviation are as follows:
Mean Deviation is the mean of all the absolute deviations of a set of data.
Quartile deviation is the difference between “first and third quartiles” in any distribution.
Standard deviation measures the “dispersion of the data set” that is relative to its mean.
Mean Deviation = 4/5 × Quartile deviation
Standard Deviation = 3/2 × Quartile deviation

To find the mode, we count how many times each number appears:

• 0 appears 3 times.

• 1 appears 1 time.

• 6 appears 5 times.

• 7 appears 2 times.

• 2 appears 2 times.

• 3 appears 1 time.

• 5 appears 1 time.

Since 6 occurs the most frequently (5 times), the mode is 6.

Thus, the correct answer is:

D: 6

Answer: A: |r| ≤ 1

Explanation:

The correlation coefficient (r) measures the strength and direction of the linear relationship between two variables. Its value always lies within the range -1 to +1, inclusive. This means:

• If r = 1, there is a perfect positive linear relationship.

• If r = 0, there is no linear relationship.

• If r = -1, there is a perfect negative linear relationship.

Thus, the absolute value of r (|r|) satisfies |r| ≤ 1. This ensures that the correlation coefficient is always bounded within the interval [-1, 1].

Why other options are incorrect:

• Option B (0 < r < 1): Incorrect because r can also take negative values (for example, -1 < r < 0).

• Option C (|r| > 1): Incorrect because the correlation coefficient cannot exceed 1 in magnitude.

• Option D (-1 < r < 0): Incorrect because r can also take positive values or be 0 or ±1.

Final Answer: A: |r| ≤ 1

Answer: B: 6

Explanation: When a constant is added to every observation in a data set, the spread or dispersion of the data does not change. Therefore, adding 1 to each observation leaves the standard deviation unchanged.

First 50 even natural numbers are 2,4,6,......,100

Mean of the scores = 202/15
Mean of the correct scores = 200/15
i.e., Mean changes.
Median is same for both cases i.e., 14.
Mode is proportional to mean.

Explanation: The total weight of the first group is 10 × 28 = 280. Let the number of items in the second group be n, so its total weight is 35 × n = 35n. For the combined group of (10 + n) items with a mean of 30, the total weight is 30 × (10 + n). Setting up the equation:

280 + 35n = 30 × (10 + n)

Expanding the right side gives:

280 + 35n = 300 + 30n

Subtracting 30n from both sides:

280 + 5n = 300

Subtracting 280 from both sides:

5n = 20

Dividing by 5:

n = 4

Now we will replace x₂ by λ
So number of elements in series will not change.
New series will include λ and exclude x₂
Hence new series sum:

Answer: D: quartile deviation

Explanation: Quartile deviation, also known as the semi-interquartile range, is based on the quartiles of the data and is less influenced by extreme values. Unlike measures such as range, standard deviation, or mean deviation, quartile deviation focuses on the middle 50% of the data, making it a more robust measure of dispersion when extreme items are present.

We have,


Now, consider

The new observations are obtained by adding 20 to each. Hence, σ does not change.

Correlation coefficient = cov (x,y)/ (std deviation (x) ×std deviation (y))
Correlation coefficient  = 0.28
cov (x,y) = 7.6
variance of x is 9.  
=> std deviation (x) = √variance  of X = √9 = 3
=>  0.28  = 7.6 / ( 3 * std deviation (y))
=> std deviation (y) = 7.6 / ( 3 * 0.28)
=> std deviation (y) = 9.05
standard deviation of Y series = 9.05

a being the meanof the whole group
Let m₂= mean of the second group

To calculate the quartile deviation (Q.D.), follow these steps:

1. Arrange the data in ascending order: 7, 10, 12, 15, 17, 19, 25

2. Find the median (middle value). Since there are 7 data points, the median is the 4th value: Median = 15

3. Divide the data into the lower and upper halves (excluding the median):

   • Lower half: 7, 10, 12

   • Upper half: 17, 19, 25

4. Find the first quartile (Q1) which is the median of the lower half: Q1 = 10

5. Find the third quartile (Q3) which is the median of the upper half: Q3 = 19

6. Compute the quartile deviation using: Quartile Deviation = (Q3 - Q1) / 2 = (19 - 10) / 2 = 9 / 2 = 4.5

Thus, the Q.D. of the daily wages is 4.5, and the correct answer is:

A: 4.5

∴ M = 55
So the class intervals can be
10−20,20−40,40−70,70−80,80−100​​

Then, variance of the height of six students

Median = [mode + 2(mean)]/3
= [6μ+2(9μ)]/3
= 24μ/3
= 8μ

Line of regression of y on x is: