Answer:
![\mu_{x}=np](https://tex.z-dn.net/?f=%5Cmu_%7Bx%7D%3Dnp)
![\sigma^2_{X}=np(1-p)](https://tex.z-dn.net/?f=%5Csigma%5E2_%7BX%7D%3Dnp%281-p%29)
Step-by-step explanation:
∵ When x is a random variable having distribution B(n, p), then for sufficiently large value of n, the following random variable has a standard normal distribution,
![\frac{x-\mu}{\sigma}\sim N(0,1)](https://tex.z-dn.net/?f=%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Csim%20N%280%2C1%29)
Where,
![\mu=np](https://tex.z-dn.net/?f=%5Cmu%3Dnp)
Here the variable X has a binomial distribution,
Such that, np (1 - p) ≥ 10 ⇒ n is sufficiently large.
Where, n is the total numbers of trials, p is success in each trials,
So, the mean of variable X is,
![\mu_{X} = np](https://tex.z-dn.net/?f=%5Cmu_%7BX%7D%20%3D%20np)
And, variance of variable X is,
![\sigma^2{X}=np(1-p)](https://tex.z-dn.net/?f=%5Csigma%5E2%7BX%7D%3Dnp%281-p%29)