Question

Olive weights are classified according to a unique set of adjectives implying great size. For example, the mean weight of olives classified as "Colossal" is 7.7 grams. Suppose a particular company’s crop of "Colossal" olives is approximately Normally distributed with a mean of 7.7 grams and a standard deviation of 0.2 grams. Which of the following represents the probability that the mean weight of a random sample of 3 olives from this population is greater than 8 grams?
a. 0.0970
b. 0.9953
c. 0.0668
d. 0.0047
e. 0.1932

Answer 1

Answer:

d. 0.0047

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

Population:

We have that $\mu = 7.7, \sigma = 0.2$

Sample of 3:

$n = 3, s = \frac{0.2}{\sqrt{3}} = 0.1155$

Which of the following represents the probability that the mean weight of a random sample of 3 olives from this population is greater than 8 grams?

This is 1 subtracted by the pvalue of Z when X = 8. So

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

By the Central Limit Theorem

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

$Z = \frac{8 - 7.7}{0.1155}$

$Z = 2.6$

$Z = 2.6$ has a pvalue of 0.9953

1 - 0.9953 = 0.0047

The probability is given by option d.