Recent studies indicate that the typical 50-year-old woman spends $350 per year for personal-care products. The distribution of the amounts spent follows a normal distribution with a standard deviation of $45 per year. We select a random sample of 40 women. The mean amount spent for those sampled is $335.

P ( x_bar > 335 ) = 0.9826

Step-by-step explanation:

Given:

- Mean amount u = 350

- standard deviation s.d = 45/year

- Sample size n = 40

Find:

- The probability of sample mean P( x_bar > 335 )

Solution:

- P ( x_bar > 335 ) = P ( Z > sqrt(n)*(x_bar - u)/s.d)

                             = P ( Z > sqrt(40)*(335-350)/45)

                             = P ( Z > -2.111) = P ( Z < 2.111)

                             =  0.5 + P( 0 < Z < 2.111)

                              = 0.5 + 0.4826

                             = 0.9826

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