# 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

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 based are selected independently of one another. How many of these 1000 intervals do you expect to capture the corresponding value of $\mu ?$
What isthe probability that between 970 and 990 of these intervals conta the corresponding value of ? (Hint: Let
‘the number among the 1000 intervals that contain What king of random variable s 2) (Use the normal approximation to the binomial distribution
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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.
In the given situation, the experiment has 1,000 trails and the probability of success 0.98 remains same for all trails. 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 random variable Y is said to follow Binomial distribution if the mass function of Y is,
$p\left(y\right)=\left(ny\right){p}^{y}{q}^{n-y},y=0,1,2,....,n$
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 μμ:
Here, $Y\sim B\left(np,npq\right).$
The expected number of intervals that contain the population mean μ is obtained as follows:
$E\left(Y\right)=n×p$
$=1,000×0.98$
$=980$
Thus, the expected number of intervals that contain the population mean mu is 980. Normal approximation to binomial distribution:
If Y follows binomial distribution with parameters np and npq that is, $Y\sim B\left(np,npq\right).$
Then for large $n,\frac{Y-np}{\sqrt{npq}}\sim N\left(0,1\right).$
That is, $Z=\frac{Y-np}{\sqrt{npq}}$
Probability $P\left(970\le Y\le 990\right):$
To obtain the probability value $P\left(970\le Y\le 990\right),$ the normal approximation to the binomial distribution has to be used.
$P\left(970\le Y\le 990\right)=P\left(\frac{970-980}{4.427}\le \frac{Y-980}{4.427}\le \frac{990-980}{4.427}\right)$
$=P\left(-2.26\le Z\le 2.26\right)$
$=P\left(Z\le 2.26\right)-P\left(Z\le -2.26\right)$
$=0.9880-0.0119$ [Using standart normal table]
$=0.9761$
Therefore the indicated probability is 0.9761.