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**Answer**

## central limit theorem practice problems

**Example 1**

The number of days a student attends classes follows a normal distribution with mean 268 days and a standard deviation of 11 days. 70 students are taken as a sample from the population. What is the statistic’s Expected value and Standard deviation.

**Solution**

The expected value E(x)= 268

The statistic standard deviation will be given by

$TheStatisticS\mathrm{tan}dardDeviation=\frac{\delta}{\sqrt{n}}$

$=\frac{\delta}{\sqrt{n}}$

$=\frac{11}{\sqrt{70}}$

$=\frac{11}{8.3666}$

=1.31475

=1.315

Example 2

Customers standing in the queue are served by bank tellers one by one. Given that the expected service time is 2 minutes and the variance is 1. Assume that the service times for the different bank customers are independent. Find the probability that P(90<Y<110) by letting Y be the total time the bank teller spends serving 50 customers

**Solution**

$Y={X}_{1}+{X}_{2}+.......+{X}_{n}$

$FromtheQuestionweknowthatn=50,E\left({X}_{i}\right)=\mu =2andthevarianceis1$

Using the concept of Z-calculations We can write;

$P(90<Y\le 110)=P(\frac{90-n\mu}{\sqrt{n\delta}}<\frac{Y-n\mu}{\sqrt{n\delta}}<\frac{110-n\mu}{\sqrt{n\delta}})$

$=P(\frac{90-50\left(2\right)}{\sqrt{50\left(1\right)}}<\frac{Y-n\mu}{\sqrt{n\delta}}<\frac{110-50\left(2\right)}{\sqrt{50\left(1\right)}})$

$=P(\frac{90-100}{\sqrt{50}}<\frac{Y-n\mu}{\sqrt{n\delta}}<\frac{110-100}{\sqrt{50}})$

$=P(-\sqrt{2}<\frac{Y-n\mu}{\sqrt{n\delta}}<\sqrt{2})$

Following the requirements of CLT, we know that $\frac{Y-n\mu}{\sqrt{n\delta}}$is tentatively standard normal.

Therefore, the equation will be rewritten as follows;

$P(90<Y\le 110)=\varphi \left(\sqrt{2}\right)-\varphi (-\sqrt{2})$

=0.8427

Example 3

In a toy production company, the company can produce 1000 toys which would have defects occuring with a probability of 0.1. It is presumed that the defects occurs independently. Find the probability that there are more than 120 defects in a given set of 1000 toys.

**Solution**

$Y={X}_{1}+{X}_{2}+.......+{X}_{n}$

The distribution defined above follows a bernoullis distribution

Click Here to see the calculation of Bernoulli Mean

Thus the mean will be p=0.1

The variance of Bernoulli distribution is derived as Click Here

Thus Variance= p(1-p)

=0.09

Using the CLT we have;

$P(Y>120)=P(\frac{Y-n\mu}{\sqrt{n\delta}}>\frac{120-n\mu}{\sqrt{n\delta}})$

$=P(\frac{Y-n\mu}{\sqrt{n\delta}}\frac{120-1000(0.1)}{\sqrt{1000(0.09)}})$

$P(Y>120)=P(\frac{Y-n\mu}{\sqrt{n\delta}}>\frac{120-100}{\sqrt{90}})$

$=1-\varphi \left(\frac{20}{\sqrt{90}}\right)$

=0.0175

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