Central limit theorem says that mean of a sampling distribution will be near normal if the sample size is at least ten percent of the total population. If the sampling population increases, the mean will follow the normal distribution. To find the probability of the mean to be greater than or less than or between certain values, find the z-scores and look up the z-scores in the z-table.

## Word Problems

**Problem 1:**

The average GPA scored by a a class is $4.91$ and standard deviation is $0.72$. For a sample of $20$ students, find the z-score that the average is above $5$.

Solution:

Average, $\mu$ = $4.91$

Standard deviation, $\sigma$ = $0.72$

Sample size, $n$ = $20$

Z-score = $\frac{X-\mu }{\frac{\sigma }{\sqrt{n}}}$

= $\frac{5-4.91 }{\frac{0.72 }{\sqrt{20}}}$ = $0.559$

Hence, the Z-score is $0.559$.**Problem 2:**

The mean salary of all employees in a company is $3578$, and standard deviation is $1980$. Find the z-score for the mean of a sample of $18$ employees to be less than $3000$.

**Solution: **

Average, $\mu$ = $3578$

Standard deviation, $\sigma$ = $1980$

Sample size, $n$ = $18$

Z-score = $\frac{X-\mu }{\frac{\sigma }{\sqrt{n}}}$

= $\frac{3000-3578}{\frac{1980}{\sqrt{18}}}$ = $1.2385$

Hence, the Z-score is $1.2385$.**Problem 3: **

The average score of a subject is $2.89$ for the whole class, with a standard deviation of $0.63$. If a sample of $25$ students is being taken, then find the probability of getting the average of this sample to be more than $3$.

**Solution: **

Average, $\mu$ = $2.89$

Standard deviation, $\sigma$ = $0.63$

Sample size, $n$ = $25$

Z-score = $\frac{X-\mu }{\frac{\sigma }{\sqrt{n}}}$

= $\frac{3-2.89}{\frac{0.63}{\sqrt{25}}}$ = $0.126$

Hence, the Z-score is $0.126$

Looking the z-score in normal curve table, the probability is found to be $0.8078$.

Hence, probability = $1$ - $0.8078$ = $0.1922$.