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
[tex]Z_{1} =\frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{24.50-25}{\frac{2.30}{\sqrt{40} } } = -1.375[/tex]
Given X₂ = 25.50
[tex]Z_{2} =\frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{25.50-25}{\frac{2.30}{\sqrt{40} } } = 1.3751[/tex]
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
