Question

A particular variety of watermelon weighs on average 22.4 pounds with a standard deviation of 1.36 pounds. Consider the sample mean weight of 64 watermelons of this variety. Assume the individual watermelon weights are independent.
Required:
a. What is the expected value of the sample mean weight?
b. What is the standard deviation of the sample mean weight?
c. What is the approximate probability the sample mean weight will be less than 22.02?
d. What is the value c such that the approximate probability the sample mean will be less than c is 0.9?

Answer 1

Answer:

a) 22.4 pounds.

b) 0.17 pounds.

c) 0.0127 = 1.27% approximate probability the sample mean weight will be less than 22.02.

d) c = 22.62

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Average 22.4 pounds with a standard deviation of 1.36 pounds.

This means that $\mu = 22.4, \sigma = 1.36$

Consider the sample mean weight of 64 watermelons of this variety.

This means that $n = 64, s = \frac{1.36}{\sqrt{64}} = 0.17$

a. What is the expected value of the sample mean weight?

By the Central Limit Theorem, 22.4 pounds.

b. What is the standard deviation of the sample mean weight?

By the Central Limit Theorem, 0.17 pounds.

c. What is the approximate probability the sample mean weight will be less than 22.02?

This is the p-value of Z when X = 22.02. So

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

By the Central Limit Theorem

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

$Z = \frac{22.02 - 22.4}{0.17}$

$Z = -2.235$

$Z = -2.235$ has a p-value of 0.0127.

0.0127 = 1.27% approximate probability the sample mean weight will be less than 22.02.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.9?

This is the 90th percentile, that is, $X = c$ when z has a p-value of 0.9, so X when Z = 1.28.

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

$1.28 = \frac{c - 22.4}{0.17}$

$c - 22.4 = 1.28*0.17$

$c = 22.62$