Question

Suppose babies born in a large hospital have a mean weight of 3242 grams, and a standard deviation of 446 grams. If 107 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 40 grams

Answer 1

64.76% probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams

Explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 3242, \sigma = 446, n = 107, s = (446)/(√(107)) = 43.12`

What is the probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams

This is the pvalue of Z when X = 3242 + 40 = 3282 subtracted by the pvalue of Z when 3242 - 40 = 3202

X = 3282

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (3282 - 3242)/(43.12)`

`Z = 0.93`

`Z = 0.93` has a pvalue of 0.8238.

X = 3202

`Z = (X - \mu)/(s)`

`Z = (3202 - 3242)/(43.12)`

`Z = -0.93`

`Z = -0.93` has a pvalue of 0.1762

0.8238 - 0.1762 = 0.6476

64.76% probability that the mean weight of the sample babies would differ from the population mean by less than 40 grams