MATH SOLVE

3 months ago

Q:
# Suppose babies born in a large hospital have a mean weight of 3215 grams, and a variance of 84,681. if 67 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams? round your answer to four decimal places.

Accepted Solution

A:

Answer:0.8558 = 85.58% probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams.Step-by-step explanation:To solve this question, we need to understand the normal probability distribution and the central limit theorem.Normal Probability DistributionProblems of normal distributions can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.Central Limit TheoremThe Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.Mean weight of 3215 grams, and a variance of 84,681This means that [tex]\mu = 3215, \sigma = \sqrt{84681} = 291[/tex]67 babies are sampled at random from the hospitalThis means that [tex]n = 67, s = \frac{291}{\sqrt{67}}[/tex]What is the probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams?p-value of Z when X = 3215 + 52 = 3267 subtracted by the p-value of Z when X = 3215 - 52 = 3163. SoX = 3267[tex]Z = \frac{X - \mu}{\sigma}[/tex]By the Central Limit Theorem[tex]Z = \frac{X - \mu}{s}[/tex][tex]Z = \frac{3267 - 3215}{\frac{291}{\sqrt{67}}}[/tex][tex]Z = 1.46[/tex][tex]Z = 1.46[/tex] has a p-value of 0.9279X = 3163[tex]Z = \frac{X - \mu}{s}[/tex][tex]Z = \frac{3163 - 3215}{\frac{291}{\sqrt{67}}}[/tex][tex]Z = -1.46[/tex][tex]Z = -1.46[/tex] has a p-value of 0.07210.9279 - 0.0721 = 0.85580.8558 = 85.58% probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams.