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4 years ago

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a. Suppose a BMW dealer in Fullerton, CA is trying to calculate the probability of his car sale for next week. The dealer knows

that the sale of car is normally distributed with mean 50 and variance 9. The variance 9 was calculated from the weekly car sale data of 20 weeks, as the population variance is not known to the dealer. What is the probability that the dealer will sell 51 or more cars next week? (Hint: use t distribution) (15)

1 answer:

egoroff_w [7]4 years ago

8 0

Answer:

0.45576

Step-by-step explanation:

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

Standard Deviation = √variance

Mean = 50

= √9

= 3

z = 51 - 50/3

= 0.11111

Probability value from Z-Table:

P(x<51) = 0.54424

P(x>51) = 1 - P(x<51)

= 1 - 0.54424

= 0.45576

The probability that the dealer will sell 51 or more cars is 0.45576

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See attachment for graph

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