Question

The suggested retail price for a certain video game is $59.24. The data below contains a random sample of 200 retail prices paid for this video game.

a. sample mean=58.53

b. std deviation=2.53

c. Determine if the average retail price has fallen from ​$59.24 by calculating the probability that a random sample of size 200 would result in a sample mean no larger than the sample mean calculated in part a. Assume that the population mean is ​$59.24 and that the sample standard deviation is representative of the population standard deviation.

The probability of obtaining a sample with a mean no larger than the sample mean calculated previously is (ans)

Hey! I'm not sure how to do c.Thanks!

Answer 1

Solution:

Here, we have to find P(Xbar < 58.53)

Z = (Xbar - µ)/[σ/sqrt(n)]

We are given

Xbar = 58.53

µ = 59.24

σ = S = 2.53

n = 200

Z = (58.53 – 59.24)/[2.53/sqrt(200)]

Z = -0.71/ 0.178898

Z = -3.96874

P(Z<-3.96874) = P(Xbar < 58.53) = 0.0000361265

(by using z-table or excel)

Required probability = 0.0000361265