Statistical Theoretical Distribution

Last Updated on: 16th June 2025, 12:06 pm

Statistical Theoretical Distribution – MCQ

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1. The mean of binomial distribution with parameter n and p is
(a) np
(b) n(1 – p)
(c) np(1 – p)
(d) $\displaystyle \sqrt{{}}$ np(1 – p) .

(a) np – Correct; mean = number of trials × probability of success.

(b) n(1 – p) – Mean of failures, not successes.

(d) √np(1 – p) – Same as (c), incorrect for mean ….

Correct option is (a)

2. In Binomial Distribution ‘p’ denotes Probability of
(a) Success
(b) Failure
(c) Both
(d) None

(a) Success – Correct; by definition.

(b) Failure – That would be q = 1 – p.

(d) None – Incorrect

Correct option is (a)

3. Standard deviation of binomial distribution is
(a) (npq)²
(b) $\displaystyle \sqrt{{npq}}$
(c) (np)²
(d) $\displaystyle \sqrt{{np}}$

(a) (npq)² – Incorrect; this gives variance squared.

(b) √(npq) – Correct formula for standard deviation.

(d) √np – Represents the mean, not standard deviation.

Correct option is (b)

4. Poisson distribution standard deviation is equal to

(a) $\displaystyle \frac{{{{e}^{m}}{{m}^{r}}}}{{r!}}\times {{r}^{2}}$

(b) $\displaystyle \frac{{{{e}^{m}}{{m}^{r}}}}{{r!}}$

(d) $\displaystyle {{e}^{{-m}}}\frac{{{{m}^{r}}}}{{r!}}$

Poisson distribution standard deviation is equal to the square-root of the mean=√m

Correct option is (d)

5. In Standard Normal distribution
(a) mean = 1, S.D. = 0
(b) mean = 1, S.D. = 1
(c) mean = 0, S.D. = 1
(d) mean = 0, S.D. = 0.

(a) Mean = 1, S.D. = 0 – Incorrect.

(b) Mean = 1, S.D. = 1 – Incorrect; mean is not 1.

(d) Mean = 0, S.D. = 0 – Would not be a distribution.

Correct option is (c)

6. In Normal distribution men, median and mode are
(a) equal
(b) not equal
(c) zero
(d) None

(a) Equal – Correct; all central values coincide in symmetric normal curve.

(b) Not equal – False.

(d) None – Incorrect.

Correct option is (a)

7. The mean of Binomial distribution is 4 and its variance is 2.4, the value of p is
(a) 0.4
(b) 0.5
(c) 0.3
(d) 0.2
np = .4
npq = 2.4
npq = $\displaystyle \frac{{2.4}}{4}$ Or, q = .60

So, P = 1 – .60 = .40

Correct option is (a)

8. For a Binomial distribution, mean = 20 and S.D. = 2 then the value of q is
(a) 0.4
(b) 0.2
(c) 0.3
(d) 0.5
np = 20
$\displaystyle \sqrt{{npq}}=2$
So, $\displaystyle {{\left( {\sqrt{{npq}}} \right)}^{2}}$ = 2² = 4
Or, npq = 4

$\displaystyle \frac{{npq}}{{np}}=\frac{4}{{20}}$

Or, q = .20

Correct option is (b)

9. For Binomial distribution, mean = 20 and S.D. = 2 then parameter n is
(a) 20
(b) 24
(c) 30
(d) 25
np = 20 $\displaystyle {\sqrt{{npq}}}$ = 2
npq = 4

$\displaystyle \frac{{npq}}{{np}}=\frac{4}{{20}}$

Or, q = .20
So, P = 1 – q = 1 – .20 = .80
Or, np = 20
or, n = $\displaystyle \frac{{20}}{P}=\frac{{20}}{{.80}}$ = 25

Correct option is (d)

10. The mean of a Poisson variate is 0.81, then its S.D. is
(a) 0.81
(b) 0.9
(c) 0.8647
(d) None
Arithmetic mean of poisson
Variate is np
np = .81
Standard deviation of poisson
Distribution = $\displaystyle {\sqrt{{np}}}$
= $\displaystyle {\sqrt{{.81}}}$ = .9

Correct option is (b)

11. x is a Poisson variate such that P(x = 3) = P(x = 4), mean =
(a) 2
(b) 3
(c) 4
(d) None
From the given information
P (x = 3) = P (x = 4)

i.e. $\displaystyle \frac{{{{e}^{{-m}}}.{{m}^{3}}}}{{3!}}=\frac{{{{e}^{{-m}}}.{{m}^{4}}}}{{4!}}$

Or $\displaystyle \frac{{{{m}^{3}}}}{{1\times 2\times 3\times }}=\frac{{{{m}^{4}}}}{{1\times 2\times 3\times 4\times }}$

Or, $\displaystyle {{m}^{3}}=\frac{{{{m}^{4}}}}{4}$

Or, m = 4

Correct option is (c)

12. For a Poisson variate x its P(x = 1) = P(x = 2), Variance is
(a) 2
(b) 3
(c) 1
(d) None
From given information
P (x = 1) = P (x = 2)

i.e. $\displaystyle \frac{{{{e}^{{-m}}}.{{m}^{1}}}}{{1!}}=\frac{{{{e}^{{-m}}}.{{m}^{2}}}}{{2!}}$

Or, m = 2
In case of poisson distribution
Mean = np
Variance = np
So, Variance = 2

Correct option is (a)

13. The probability mass function of binomial distribution is given by
(a) f(x) = p ^x q ^n-x
(b) f(x) = ⁿC_x p^x q ^n-x
(c) f(x) = ⁿC_x q^x p ^n-x
(d) f(x) = ⁿC_x p^n-x q^x

(a) f(x) = p ^x q ^n-x not correct – Lacks ⁿC_x

(b) f(x) = ⁿC_x p^x q ^n-x – Complete and correct.

(d) f(x) = ⁿC_x p^n-xq^x – Also incorrect order.

Correct option is (b)

14. The maximum value of the variance of a binomial distribution with parameters n and p is
(a) n/2
(b) n/4
(c) np (1-p)
(d) 2n.

The variance of a binomial distribution is given by the formula: Variance = np(1-p), where :

n: The number of trials, which is a constant in a given binomial distribution.

p: The probability of success in a single trial.

To maximize the product np(1-p), we can consider the function f(p) = p(1-p).

f(p) = p(1-p) = p – p²

f'(p) = 1 – 2p

Setting f'(p) = 0, we get 1 – 2p = 0, or p = 1/2

Therefore, the maximum value of the variance is achieved when p = 1/2. Substituting p = 1/2 into the variance formula:

Variance = n X (1/2) X (1 – 1/2) = n X (1/2) X (1/2) = n/4

Correct option is (b)

15. What is the probability of getting 3 heads if 6 unbiased coins are tossed simultaneously?
(a) 0.50
(b) 0.25
(c) 0.3125
(d) 0.6875

Correct option is (c)

16. If for a Poisson variable X, f(2) = 3 f(4), what is the variance of X?
(a) 2
(b) 4
(c) $\displaystyle \sqrt{2}$
(d) 3

Poisson: f(2) = 3×f(4)

So, m²/2! = 3 × m⁴/4!. Or, m² = m⁴/4, or m² = 4, Or, m = 2

For Poisson distribution, mean = variance,

So, when m = 2, Variance = m = 2

Correct option is (a)

17. If the mean deviation of a normal variable is 16, what is its quartile deviation?
(a) 10.00
(b) 13.30
(c) 15.00
(d) 12.05
For normal distribution

M.D. = $\displaystyle \frac{4}{5}$ . S.D.