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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 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. What is the likelihood of finding a sample mean this large or larger from the specified population

Respuesta :

Answer:

The likelihood is 0.9826.

Step-by-step explanation:

Here mean amount or μ = 350

Standard deviation = 45

Sample size n = 40

Let X be the sample mean where X>335

P(X>335)=[tex]P(z>\frac{335-350}{45/\sqrt{40} } )[/tex]

[tex]P(z>\frac{-15}{7.12})[/tex]

[tex]P(z>-2.11)[/tex]

[tex]P(z<2.11)[/tex]

=>[tex]0.5+P(0<z<2.11)[/tex]

=> [tex]0.5+0.4826=0.9826[/tex]

Hence, the likelihood is 0.9826.

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