Answer:
The probability that the average final exam grade of this sample is between 77 and 82
P(77≤ x⁻≤ 82 = 0.8315 or 83%
Step-by-step explanation:
Step(i):-
Given random sample size 'n' = 40
Mean of the normal distribution = 81
Standard deviation of normal distribution = 6.6
Let x₁⁻ = 77
[tex]Z_{1} = \frac{x_{1} -mean }{\frac{S.D}{\sqrt{n} } } = \frac{77-81}{\frac{6.6}{\sqrt{40} } }[/tex]
Z₁ = -3.83
Let x₂⁻ = 77
[tex]Z_{2} = \frac{x^{-} _{2} -mean }{\frac{S.D}{\sqrt{n} } } = \frac{82-81}{\frac{6.6}{\sqrt{40} } }[/tex]
Z₂ = 0.958
The probability that the average final exam grade of this sample is between 77 and 82
P(77≤ x≤ 82) = P( -3.83 ≤x≤0.958)
= A( 0.958) + A(3.83)
= 0.3315 + 0.4995
= 0.8315
The probability that the average final exam grade of this sample is between 77 and 82 = 0.8315 or 83%