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Two stores sell watermelons. At the first store, the melons weigh an average of 22 pounds, with a standard deviation of 2.5 pounds. At the second store, the melons are smaller, with a mean of 18 pounds and a standard deviation of 2 pounds. You select a melon at random at each store.

  1. Find the mean difference in weights of the melons?
  2. Find the standard deviation of the difference in weights?
  3. If a Normal model can be used to describe the difference in weights, find the probability that the melon you got at the first store is heavier?

This question aims to find the mean difference and standard deviation in the difference in weights of the melons from two stores. Also, to check if the melon from the first store is heavier.

The question is based on the concepts of probability from a normal distribution using a z-table or z-score. It also depends on the population’s mean and the population’s standard deviation. The z-score is the deviation of a data point from the population’s mean. The formula for z-score is given as:

\[ z = \dfrac{ x\ -\ \mu}{ \sigma } \]

Expert Answer

The given information about this problem is as follows:

\[ Mean\ Weight\ of\ Melons\ from\ First\ Store\ \mu_1 = 22 \]

\[ Standard\ Deviation\ of\ Weight\ of\ Melons\ from\ First\ Store\ \sigma_1 = 2.5 \]

\[ Mean\ Weight\ of\ Melons\ from\ Second\ Store\ \mu_2 = 18 \]

\[ Standard\ Deviation\ of\ Weight\ of\ Melons\ from\ Second\ Store\ \sigma_2 = 2 \]

a) To calculate the mean difference between weights of the melons from the first and second store, we simply need to take the difference of the means of both stores. The mean difference is given as:

\[ \mu = \mu_1\ -\ \mu_2 \]

\[ \mu = 22\ -\ 18 \]

\[ \mu = 4 \]

b) To calculate the standard deviation in difference in weights of the melons from both stores, we can use the following formula which is given as:

\[ SD = \sqrt{ \sigma_1^2 + \sigma_2^2 } \]

Substituting the values, we get:

\[ SD = \sqrt{ 2.5^2 + 2^2 } \]

\[ SD = \sqrt{ 6.25 + 4 } \]

\[ SD = \sqrt{ 10.25 } \]

\[ SD = 3.2016 \]

c) The normal model of the differences in mean and standard deviation can be used to calculate the probability that the melon from the first store is heavier than the melon from the second store. The formula to calculate z-score is given as:

\[ z = \dfrac{ x\ -\ \mu}{ \sigma } \]

Substituting the values, we get:

\[ z = \dfrac{ 0\ -\ 4 }{ 3.2016 } \]

\[ z = -1.25 \]

Now we can calculate the probability using the z-table.

\[ P(Z \gt 1.25) = 1\ -\ P(Z \lt -1.25) \]

\[ P(Z \gt 1.25) = 1\ -\ 0.1056 \]

\[ P(Z \gt 1.25) = 0.8944 \]

Numerical Result

a) The mean difference in the weights of the melons between the first and second store is calculated to be 4.

b) The standard deviation of the difference in weights is calculated to be 3.2016.

c) The probability that the melon from the first is heavier than the melon from the second store is calculated to be 0.8944 or 89.44%.

Example

The mean of a sample is given as 3.4 and the standard deviation of the sample is given as 0.3. Find the z-score of a random sample of 2.9.

The formula for z-score is given as:

\[ z = \dfrac{ x\ -\ \mu}{ \sigma } \]

Substituting the values, we get:

\[ z = \dfrac{ 2.9\ -\ 3.4 }{ 0.3 } \]

\[ z = -1.67 \]

The probability associated with this z-score is given as 95.25%.

5/5 - (16 votes)