Respuesta :
Answer:
The probability that the mean of a sample of 36 cars would be less than 3245 miles
P(X⁻≤3245) = 0.975
Step-by-step explanation:
Step(i):-
The mean number of miles between services
μ= 3408 miles
The Variance of miles between services
σ² = 249,001
σ = √249,001 = 499
Let 'X' be a random variable in a normal distribution
Given sample size 'n' =36
Step(ii):-
[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]
Z = -163/83.16 = 1.96
The probability that the mean of a sample of 36 cars would be less than 3245 miles
P(X⁻≤3245) = P(Z≤1.96)
= 0.5 + A(1.96)
= 0.5 + 0.4750
= 0.975
Final answer:-
The probability that the mean of a sample of 36 cars would be less than 3245 miles
P(X⁻≤3245) =0.975
