Using the binomial distribution, find the mean and variance of X for the following experiments. (i) A fair coi

Exercise 11.5 · Q3

X

(i) A fair coin is tossed

100

times, and

X

denotes the number of heads.

(ii) A fair die is tossed

240

times, and

X

denotes the number of times that four appeared.

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✓ Free question

Concept understanding — Mean & Variance of Standard Distributions

Four named discrete distributions, in order of increasing generality, each with its own closed-form mean and variance:

One-point distribution ( $f(x_0)=1$ ): mean $=x_0$ , variance $=0$ — no spread is possible when all mass sits at one value (matches Corollary 4 of the expectation properties, $V(b)=0$ ).
Two-point distribution (values $x_1,x_2$ with probabilities $p,q=1-p$ ): mean $=px_1+qx_2$ , variance $=pq(x_2-x_1)^2$ ; in the symmetric case $p=q=\tfrac12$ these become $\dfrac{x_1+x_2}{2}$ and $\dfrac{(x_2-x_1)^2}{4}$ .
Bernoulli distribution $X\sim\text{Ber}(p)$ (the two-point distribution specialised to $x_1=1,x_2=0$ ): mean $=p$ , variance $=pq$ ; symmetric case ( $p=\tfrac12$ ) gives mean $\tfrac12$ , variance $\tfrac14$ .
Binomial distribution $X\sim B(n,p)$ (the sum of $n$ independent Bernoulli $(p)$ trials): mean $=np$ , variance $=np(1-p)=npq$ ; symmetric case ( $p=\tfrac12$ ) gives mean $\dfrac n2$ , variance $\dfrac n4$ .

Reading a problem "backwards" from mean/variance. A large family of exercises states $\mu=np$ and $\sigma^2=npq$ (or $\sigma$ ) numerically and asks for $n,p$ (and hence the full distribution) — divide the two given equations to eliminate $n$ : $\dfrac{\sigma^2}{\mu}=\dfrac{npq}{np}=q$ , giving $q$ (hence $p=1-q$ ) directly; then $n=\mu/p$ . The same pair of equations, run the other way, instantly gives the standard deviation or variance once $n,p$ are known ( $\sigma=\sqrt{np(1-p)}$ ).

Sanity checks. $0<p<1\Rightarrow0<q<1$ , and $n$ must come out a positive integer — if solving the mean/variance pair yields a non-integer or negative $n$ , recheck the algebra (a sign or ratio slip is the usual cause, since $q=\sigma^2/\mu$ must itself lie strictly between $0$ and $1$ ).

Both parts are direct binomial mean/variance substitutions once $n,p$ are identified from the experiment.

Part (i): fair coin tossed $100$ times, $X=$ number of heads. $n=100,\ p=\dfrac12,\ q=\dfrac12$ .

Mean $=np=100\times\dfrac12=50$ .

Variance $=npq=100\times\dfrac12\times\dfrac12=25$ .

Part (ii): fair die tossed $240$ times, $X=$ number of times $4$ appears. $n=240,\ p=\dfrac16,\ q=\dfrac56$ .

Mean $=np=240\times\dfrac16=40$ .

Variance $=npq=240\times\dfrac16\times\dfrac56=\dfrac{240\times5}{36}=\dfrac{1200}{36}=\dfrac{100}3\approx33.33$ .

✓Final answer

=50

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 markMCQ
$n=25$
(a) $3$
(b) $6$
(c) $2$
(d) $4$
›Reveal solutionSolution
Applies the binomial variance formula $np(1-p)$ and takes its square root for the standard deviation.

For $X\sim\text{Binomial}(n,p)$ , variance $=np(1-p)$ .

Here $n=25$ , $p=0.8$ , so $1-p=0.2$ : $\text{Var}(X)=25\times0.8\times0.2=25\times0.16=4$ .

Standard deviation $=\sqrt{\text{Var}(X)}=\sqrt4=2$ .

✓Final answer
$2$
CBSE 2020Set ANNUAL1 markMCQ
$n=25$
(a) $2$
(b) $6$
(c) $4$
(d) $3$
›Reveal solutionSolution
For $X\sim B(25,0.8)$ , the variance $npq=4$ , so the standard deviation is $\sqrt4=2$ .

For a binomial random variable $X$ with parameters $n$ (number of trials) and $p$ (probability of success), the variance is $\mathrm{Var}(X)=npq$ , where $q=1-p$ .

Here $n=25$ and $p=0.8$ , so $q=1-0.8=0.2$ .

Compute the variance: $\mathrm{Var}(X)=npq=25\times0.8\times0.2$ .

$25\times0.8=20$ , and $20\times0.2=4$ . So $\mathrm{Var}(X)=4$ .

The standard deviation is the square root of the variance: $\sigma=\sqrt{\mathrm{Var}(X)}=\sqrt4=2$ .

✓Final answer
$X$
CBSE 2018Set ANNUAL1 markMCQ
$n$
(a) $\left(\dfrac{1}{5}, 25\right)$
(b) $\left(\dfrac{4}{5}, 25\right)$
(c) $\left(25, \dfrac{1}{5}\right)$
(d) $\left(25, \dfrac{4}{5}\right)$
›Reveal solutionSolution
From mean $=np=5$ and variance $=npq=4$ , we get $q=4/5$ , $p=1/5$ , and $n=25$ .

For a binomial distribution, mean $=np$ and variance $=npq$ where $q=1-p$ .

Given mean $=5$ : $np=5$ .

Given standard deviation $=2$ , so variance $=\sigma^2=4$ : $npq=4$ .

Divide the variance equation by the mean equation: $\dfrac{npq}{np}=q=\dfrac{4}{5}$ .

So $p=1-q=1-\dfrac45=\dfrac15$ .

From $np=5$ : $n=\dfrac{5}{p}=\dfrac{5}{1/5}=25$ .

Check: $npq=25\times\dfrac15\times\dfrac45=25\times\dfrac{4}{25}=4$ . Confirmed.

✓Final answer
$n=25,\ p=\dfrac15$
CBSE 2017Set ANNUAL1 markMCQ
$Q. The mean of a binomial distribution is 5 and its standard deviation is 2. Then the values of n and p are :$
(a) $\left(\dfrac{4}{5}, 25\right)$
(b) $\left(25, \dfrac{4}{5}\right)$
(c) $\left(\dfrac{1}{5}, 25\right)$
(d) $\left(25, \dfrac{1}{5}\right)$
›Reveal solutionSolution
From mean $np=5$ and variance $npq=4$ : $q=\dfrac45,\ p=\dfrac15,\ n=25$ .

For a binomial distribution, mean $=np=5$ and standard deviation $=2\Rightarrow$ variance $=npq=2^2=4$ .

Divide variance by mean: $\dfrac{npq}{np}=q=\dfrac{4}{5}$ .

So $p=1-q=1-\dfrac45=\dfrac15$ .

From $np=5$ : $n=\dfrac{5}{p}=\dfrac{5}{1/5}=25$ .

✓Final answer
$(n,p)=\left(25,\dfrac15\right)$
CBSE 2016Set ANNUAL1 markMCQ
$Q. In 16 throws of a die getting an even number is considered a success, then the variance of the successes is$
(a) $4$
(b) $6$
(c) $2$
(d) $256$
›Reveal solutionSolution
This is a binomial variance problem: $n=16$ , $p=q=\frac12$ , giving variance $npq=4$ .

A fair die has outcomes $\{1,2,3,4,5,6\}$ ; the even numbers $\{2,4,6\}$ are $3$ out of $6$ equally likely outcomes.

Probability of success (even number) $p=\dfrac{3}{6}=\dfrac12$ ; probability of failure $q=1-p=\dfrac12$ .

The $16$ throws are independent, identical Bernoulli trials, so the number of successes follows a Binomial distribution $B(n,p)$ with $n=16,\ p=\frac12$ .

The variance of a binomial distribution is $\text{Var}=npq$ .

Substituting: $\text{Var}=16\times\dfrac12\times\dfrac12=16\times\dfrac14=4$ .

Distractors: (b) $6$ , (c) $2$ , and (d) $256$ (which is $16^2$ , an unrelated quantity) do not follow from $npq$ with the correct $p=q=\frac12$ .

✓Final answer
$4$

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