Recent studies indicate that the typical 50-year-old woman spends $350 per year

Question

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? (Round z value to 2 decimal places and final answer to 4 decimal places.)

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.)

Explanation / Answer

population mean = 350
population sigma = 45
sample size, n = 40
sample mean = 335

here we need to find the probability of finding the mean of sample greater than or equal to 335. This we can find using central limit theorem.

P(X>335) = P(z > (335 - 350)/(45/sqrt(40))) = P(z > -2.11) = 0.9825

here to find the value of z, as per central limit theorem
z = (x - mu)/(sigma/sqrt(n))

To find the probability standard z right tailed table can be used where look for the probability associated with z = 2.11 and then substract this value from 1 to get the desired probability.

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