Question

At a local sandwich shop the average customer spends $25 at dinner with a standard deviation of $2.30. In a simple random sample of 40 customers, what is the probability that the mean receipt for dinner is between $24.50 and $25.50

Answer 1

Answer:

The probability that the mean receipt for dinner is between $24.50 and $25.50 is

P(24.50≤ x ≤ 25.50)= 0.8324

Step-by-step explanation:

Step(i):-

Given mean of Population(μ) = $25

Given standard deviation of the Population  (σ) = 2.30

Sample size 'n' = 40

Let 'X' be the random variable in normal distribution

Given X₁ = 24.50

$Z_{1} =\frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{24.50-25}{\frac{2.30}{\sqrt{40} } } = -1.375$

Given X₂ = 25.50

$Z_{2} =\frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{25.50-25}{\frac{2.30}{\sqrt{40} } } = 1.3751$

Step(ii):-

The probability that the mean receipt for dinner is between $24.50 and $25.50

P(x₁≤ x ≤ x₂) = P(Z₁≤ z ≤ Z₂) = A(Z₂) + A(z₁)

P(24.50≤ x ≤ 25.50) = P(-1.375≤ z ≤ 1.375) = A(1.375) + A(1.375)

P(24.50≤ x ≤ 25.50)= 2 A( 1.375)

                               = 2 × 0.4162 ( from normal table)

                              = 0.8324

Final answer:-

The probability that the mean receipt for dinner is between $24.50 and $25.50 is 0.8324