Consider the next 1000 98% Cis for mu that a statistical consultant will obtain for various clients. Suppose the data sets on which the intervals are

For various clients 1,000 intervals of population mean are computed at 98% confidence level. 
The data sets are selected independently to compute the confidence intervals. 
Requirements for a binomial experiment: 
The experiment consists of n independent trials. 
There are only 2 possible outcomes for each trial, success and failure. 
The probability of success (S) remains the same for all trials. 
The trials are independent of each other. 
The experiment in the case comprises 1,000 trials, and the likelihood of success is 0.98 for each trial. Further all these 1,000 confidence intervals are independent of each other. 
Thus, the random variable Y = Number of intervals that contain $\mu$ among 1,000 intervals be the Binomial random variable. 
Binomial Distribution: 
a chance factor If Y's mass function is as follows, then Y is said to follow a binomial distribution.
$p(y)=(ny)p^y{q}^{n-y},y=0,1,2,....,n$ 
Where the mean is $\mu =np$ and the variance is $\sigma 2=npq.$ 
Consider a random variable Y, such that: 
Y = Number of intervals that contain μ among 1,000 intervals 
Then, the random variable Y has a binomial distribution, where there are 1,000 trials and the probability of “success” (observing the population mean μ) is 0.98. 
Thus, $n=1,000,p=0.98$ 
Expected number of intervals with population mean μ: 
Here, $Y\sim B(np,npq).$ 
The expected number of intervals that contain the population mean μ is obtained as follows: 
$E(Y)=n\times p$ 
$=1,000\times 0.98$ 
$=980$ 
Thus, the expected number of intervals that contain the population mean $\mu$ is 980. Normal approximation to binomial distribution: 
If Y has a binomial distribution with np and npq as its parameters, then $Y\sim B(np,npq).$ 
Next, for large $n,\frac{Y-np}{\sqrt{npq}}\sim N(0,1).$ 
That is, $Z=\frac{Y-np}{\sqrt{npq}}$ 
Probability $P(970\le Y\le 990):$ 
To obtain the probability value $P(970\le Y\le 990),$ it is necessary to utilize the binomial distribution's normal approximation.
$P(970\le Y\le 990)=P(\frac{970-980}{4.427}\le \frac{Y-980}{4.427}\le \frac{990-980}{4.427})$ 
$=P(-2.26\le Z\le 2.26)$ 
$=P(Z\le 2.26)-P(Z\le -2.26)$ 
$=0.9880-0.0119$ [Using standart normal table] 
$=0.9761$ 
Therefore the indicated probability is 0.9761.