Q.A die is tossed twice. A success is getting an odd number on a toss. Find the mean and the variance of the probability distribution of the number of successes.

Applies the binomial variance formula $np(1-p)$ and takes its square root for the standard deviation.

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

For $X\sim B(25,0.8)$, the variance $npq=4$, so the standard deviation is $\sqrt4=2$.

1. 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 $q=1-p$.
2. Here $n=25$ and $p=0.8$, so $q=1-0.8=0.2$.
3. Compute the variance: $\mathrm{Var}(X)=npq=25\times0.8\times0.2$. …

From mean $=np=5$ and variance $=npq=4$, we get $q=4/5$, $p=1/5$, and $n=25$.

1. For a binomial distribution, mean $=np$ and variance $=npq$ where $q=1-p$.
2. Given mean $=5$: $np=5$.
3. Given standard deviation $=2$, so variance $=\sigma^2=4$: $npq=4$.
4. Divide the variance equation by the mean equation: $\dfrac{npq}{np}=q=\dfrac{4}{5}$.
5. So $p=1-q=1-\dfrac45=\dfrac15$. …

From mean $np=5$ and variance $npq=4$: $q=\dfrac45,\ p=\dfrac15,\ n=25$.

1. For a binomial distribution, mean $=np=5$ and standard deviation $=2\Rightarrow$ variance $=npq=2^2=4$.
2. Divide variance by mean: $\dfrac{npq}{np}=q=\dfrac{4}{5}$. …

This is a binomial variance problem: $n=16$, $p=q=\frac12$, giving variance $npq=4$.

1. 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.
2. Probability of success (even number) $p=\dfrac{3}{6}=\dfrac12$; probability of failure $q=1-p=\dfrac12$.
3. The $16$ throws are independent, identical Bernoulli trials, so the number of successes follows a Binomial distribution $B(n,p)$ with $n=16,\ p=\frac12$.
4. The variance of a binomial distribution is $\text{Var}=npq$. …