Question

Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1 ounces.
(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces? Round your answer to 4 decimal places.

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces? Round your answer to 4 decimal places.

Answer 1

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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 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.

Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that $\mu = 8, \sigma = 1.1$

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

$n = 5$ means that $s = \frac{1.1}{\sqrt{5}} = 0.4919$

This probability is the pvalue of Z when X = 9.3. So

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

By the Central Limit Theorem

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

$Z = \frac{9.3 - 8}{0.4919}$

$Z = 2.64$

$Z = 2.64$ has a pvalue of 0.9959

0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?

$n = 6$ means that $s = \frac{1.1}{\sqrt{6}} = 0.4491$

This probability is 1 subtracted by the pvalue of Z when X = 9. So

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

$Z = \frac{9 - 8}{0.4491}$

$Z = 2.23$

$Z = 2.23$ has a pvalue of 0.9871

1 - 0.9871 = 0.0129

0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces