Answer:
(a) The probability that the store’s revenues were at least $9,000 is 0.0233.
(b) The revenue of the store on the worst 1% of such days is $7,631.57.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take 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,
![\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)
It is provided that:
![\mu=\$27\\\sigma=\$18\\n=310](https://tex.z-dn.net/?f=%5Cmu%3D%5C%2427%5C%5C%5Csigma%3D%5C%2418%5C%5Cn%3D310)
As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.
(a)
Compute the probability that the store’s revenues were at least $9,000 as follows:
![P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28S%5Cgeq%209000%29%3DP%28%5Cfrac%7BS-%5Cmu_%7BX%7D%7D%7B%5Csigma_%7BX%7D%7D%5Cgeq%20%5Cfrac%7B9000-%2827%5Ctimes310%29%7D%7B%5Csqrt%7B310%7D%5Ctimes%2018%7D%29%5C%5C%5C%5C%3DP%28Z%5Cgeq%201.99%29%5C%5C%5C%5C%3D1-P%28Z%3C1.99%29%5C%5C%5C%5C%3D1-0.97670%5C%5C%5C%5C%3D0.0233)
Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.
(b)
Let <em>s</em> denote the revenue of the store on the worst 1% of such days.
Then, P (S < s) = 0.01.
The corresponding <em>z-</em>value is, -2.33.
Compute the value of <em>s</em> as follows:
![z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bs-%5Cmu_%7BX%7D%7D%7B%5Csigma_%7BX%7D%7D%5C%5C%5C%5C-2.33%3D%5Cfrac%7Bs-8370%7D%7B316.923%7D%5C%5C%5C%5Cs%3D8370-%282.33%5Ctimes%20316.923%29%5C%5C%5C%5Cs%3D7631.56941%5C%5C%5C%5Cs%5Capprox%20%5C%247%2C631.57)
Thus, the revenue of the store on the worst 1% of such days is $7,631.57.