Answer:
The probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we select appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.
Then, the mean of the distribution of the sum of values of X is given by,
![\mu_{x}=n\mu](https://tex.z-dn.net/?f=%5Cmu_%7Bx%7D%3Dn%5Cmu)
And the standard deviation of the distribution of the sum of values of X is given by,
![\sigma_{x}=\sqrt{n}\sigma](https://tex.z-dn.net/?f=%5Csigma_%7Bx%7D%3D%5Csqrt%7Bn%7D%5Csigma)
The information provided is:
<em>μ</em> = $970
<em>σ</em> = $129
<em>n</em> = 102
Since the sample size is quite large, i.e. <em>n</em> = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.
Then,
![\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)](https://tex.z-dn.net/?f=%5Csum%20X%5Csim%20N%28%5Cmu_%7Bx%7D%3D98940%2C%5C%20%5Csigma_%7Bx%7D%3D1302.84%29)
Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:
![P (\sum X \leq 100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}}](https://tex.z-dn.net/?f=P%20%28%5Csum%20X%20%5Cleq%20%20100%2C000%29%20%3DP%28%5Cfrac%7B%5Csum%20X-%5Cmu_%7Bx%7D%7D%7B%5Csigma_%7Bx%7D%7D%20%3C%5Cfrac%7B100000-98940%7D%7B1302.84%7D%29)
![=P(Z](https://tex.z-dn.net/?f=%3DP%28Z%3C0.81%29%5C%5C%3D1-0.79103%5C%5C%3D0.20897%5C%5C%5Capprox0.2090)
*Use a <em>z</em>-table for the probability.
Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.