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:

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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

Population:

We have that [tex]\mu = 7.7, \sigma = 0.2[/tex]

Sample of 3:

[tex]n = 3, s = \frac{0.2}{\sqrt{3}} = 0.1155[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{8 - 7.7}{0.1155}[/tex]

[tex]Z = 2.6[/tex]

[tex]Z = 2.6[/tex] has a pvalue of 0.9953

1 - 0.9953 = 0.0047

The probability is given by option d.

You can calculate the z score for the specified sample and then use the z tables to find the p value(probability) needed.

The probability that the mean weight of a random sample of 3 olives from this population is greater than 8 grams is given by

Option d : 0.0047

Suppose that the sample is of size 'n', then we have the z score(we are converting random variable X to standard random variable Z) as:

[tex]Z = \dfrac{X - \overline{x}}{s} = \dfrac{X - \mu}{\sigma/\sqrt{n}}[/tex]

where [tex]\overline{x}[/tex] is mean of the sample

s is the standard deviation of the sample,

and we used the central limit theorem which says that a sample from a normally distributed population with [tex]mean = \mu[/tex] and standard deviation = [tex]\sigma[/tex] can have its mean approximated by population mean and its standard deviation approximated by [tex]s = \dfrac{\sigma}{\sqrt{n}}[/tex]

Let the weight of the the olives for the given crop of olives of a particular company for taken random sample is given by X (a random variable)

Then, we have:

[tex]X \sim N(7.7, 0.2)[/tex]

where

[tex]\mu = 7.7, \sigma = 0.2[/tex]

Thus, we have:

[tex]\overline{x} = \mu, s = \sigma/\sqrt{n} = 0.2/\sqrt{3}[/tex]

Using the given facts, we get the needed probability as:

[tex]P( X> 8)[/tex] sample size n  = 3)

Then, using the z score, we get):

[tex]P(X > 8) = 1 - P(X \leq 8) = 1 - P(Z \leq \dfrac{8 - 7.7}{0.2/\sqrt{3}})\\\\P(X > 8)=1- P(Z \leq \dfrac{0.3\sqrt{3}}{0.2}) = 1- P(Z \leq 1.5 \times \sqrt{3} )\\\\P(X > 8) = 1 - P(Z \leq 2.59) = 1- 0.9952 \approx 0.0047[/tex]

Thus,

The probability that the mean weight of a random sample of 3 olives from this population is greater than 8 grams is given by

Option d : 0.0047

Learn more about standard normal distribution here:

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