Question

(i) Let $Z$ be a standard normal variate. Find the value of $c$ if $P(Z<c)=0.05$
Here $P(Z<c)=0.05$
Here $P[0<Z<1.65]=0.45$
(ii) The difference between the mean and the variance of a Binomial distribution is $1$ and the difference between their squares is $11$. Find $n$

Answer 1

$\left(i\right)$

Let $Z$ be the standard normal variate.

We have to find the value of $c$ if $P(Z<c)=0.05$

Consider $P(Z<c)=0.05$

$\Rightarrow P(-\infty <Z<c)=0.05$

As area is $<0.05$, $c$ lies to the left of $z=0$

From the area table $Z$ value for the area $0.45$ is $1.65$

Therefore $c=-1.65$

$\left(ii\right)$

Let the mean be $(m+1)$ and the variance be $m$ from the given data [since mean $>$ variance in binomial distribution]

$(m+1{)}^2-m^2=11$

$\Rightarrow m=5$

$\therefore mean=m+1=6$

$\Rightarrow np=6,npq=5$

$\therefore q=\frac{5}{6},p=\frac{1}{6}$

Since $npq=5$

Substitute $q=\frac{5}{6},p=\frac{1}{6}$ we get

$n\times \frac{1}{6}\times \frac{5}{6}=5$

$\Rightarrow n\times \frac{1}{36}=1$

$\therefore n=36$