Question

A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard deviation of 18$. Suppose the store had 310 customers this Sunday. A) estimate the probability that the store’s revenues were at least 9,000$ B) if. On a typical Sunday, the store serves 310 customers, how much does he store take in on the worst 1% of such days ?

Answer 1

(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.

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`

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

`\sigma_(X)=√(n)\sigma`

It is provided that:

`\mu=\$27\\\sigma=\$18\\n=310`

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:

`P(S\geq 9000)=P((S-\mu_(X))/(\sigma_(X))\geq (9000-(27*310))/(√(310)* 18))\\\\=P(Z\geq 1.99)\\\\=1-P(Z<1.99)\\\\=1-0.97670\\\\=0.0233`

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:

`z=(s-\mu_(X))/(\sigma_(X))\\\\-2.33=(s-8370)/(316.923)\\\\s=8370-(2.33* 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57`

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.