Respuesta :
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,
[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]
It is provided that:
[tex]\mu=\$27\\\sigma=\$18\\n=310[/tex]
As the sample size is quite large, i.e. n = 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:
[tex]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<1.99)\\\\=1-0.97670\\\\=0.0233[/tex]
Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.
(b)
Let s denote the revenue of the store on the worst 1% of such days.
Then, P (S < s) = 0.01.
The corresponding z-value is, -2.33.
Compute the value of s as follows:
[tex]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[/tex]
Thus, the revenue of the store on the worst 1% of such days is $7,631.57.
