Recent studies indicate that the typical 50-year-old woman spends $353 per year

Question

Recent studies indicate that the typical 50-year-old woman spends $353 per year for personal-care products. The distribution of the amounts spent follows a normal distribution with a standard deviation of $44 per year. We select a random sample of 46 women. The mean amount spent for those sampled is $301.

What is the likelihood of finding a sample mean this large or larger from the specified population? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Recent studies indicate that the typical 50-year-old woman spends $353 per year for personal-care products. The distribution of the amounts spent follows a normal distribution with a standard deviation of $44 per year. We select a random sample of 46 women. The mean amount spent for those sampled is $301.

Explanation / Answer

Mean = 353

SD = 44

n = 46

Here we have to find the probability of P ( x bar >= 301)

The z-score for 301 = (301 – 353 ) / ( 44 / sqrt (46))

                                    = -8.02

P (z >= -8.02)

= 1 – P ( z <-8.02)

= 1 – 0 [ By using Normal Distribution Table]

= 1

Answer: 1.0000

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