# Sample mean

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Recent studies indicate that the typical 50-year-old woman spends $350 per year for personal-care products. The distribution of the amounts spent is positively skewed. We select a random sample of 40 women. The mean amount spent for those sampled is $335, and the standard deviation of the sample is $45. What is the likelihood of finding a sample mean this large or larger from the specified population?

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The likelihood of finding a sample mean this large or larger from the specified population?

Probability i.e z= (Sample mean-Population ...

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This is a step by step instruction of the likelihood of finding a sample mean this large or larger from the specified population.

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