A small business owner estimates his mean daily profit as $970 with a standard deviation of $129. His shop is open 102 days a year. What is the probability that his annual profit will not exceed $100,000?

Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we select appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

[tex]\mu_{x}=n\mu[/tex]

And the standard deviation of the distribution of the sum of values of X is given by,

[tex]\sigma_{x}=\sqrt{n}\sigma[/tex]

The information provided is:

μ = $970

σ = $129

n = 102

Since the sample size is quite large, i.e. n = 102 > 30, the Central Limit Theorem can be used to approximate the distribution of the shopkeeper's annual profit.

Then,

[tex]\sum X\sim N(\mu_{x}=98940,\ \sigma_{x}=1302.84)[/tex]

Compute the probability that the shopkeeper's annual profit will not exceed $100,000 as follows:

[tex]P (\sum X \leq 100,000) =P(\frac{\sum X-\mu_{x}}{\sigma_{x}} <\frac{100000-98940}{1302.84})[/tex]

[tex]=P(Z<0.81)\\=1-0.79103\\=0.20897\\\approx0.2090[/tex]

*Use a z-table for the probability.

Thus, the probability that the shopkeeper's annual profit will not exceed $100,000 is 0.2090.