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 more than 40 grams? Round your answer to four decimal places.

Answer 1

Answer:

0.3530 = 35.30% probability that the mean weight of the sample babies would differ from the population mean by more than 40 grams

Step-by-step explanation:

To solve this question, we need 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:

$Z = \frac{X - \mu}{\sigma}$

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 3242, \sigma = 446, n = 107, s = \frac{446}{\sqrt{107}} = 43.12$

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 more than 40 grams?

Either it differs by 40 grams or less, or it differs by more than 40 grams. The sum of the probabilities of these events is decimal 1.

Probability it differs by 40 grams of less

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

X = 3282

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{3282 - 3242}{43.12}$

$Z = 0.93$

$Z = 0.93$ has a pvalue of 0.8238

X = 3202

$Z = \frac{X - \mu}{s}$

$Z = \frac{3202 - 3242}{43.12}$

$Z = -0.93$

$Z = -0.93$ has a pvalue of 0.1762

0.8238 - 0.1768 = 0.6470

0.6470 probability it differs by 40 grams or less

Probability it differs by more than 40 grams

p + 0.6470 = 1

p = 0.3530

0.3530 = 35.30% probability that the mean weight of the sample babies would differ from the population mean by more than 40 grams