2. Scotts American Bulldog males have weights that are normally distributed with

a mean of 85 pounds and a standard deviation of 14.6 pounds. Female American

Bulldog have weights that are normally distributed with a mean of 69 pounds and

a standard deviation of 8.8 pounds. We take a sample of 9 male and 8 female

American bulldogs. What is the probability that the difference in mean weight of

the samples (Male - Female) is greater than fifteen pounds? (Please attach your

work).

Answer:

0.5675 = 56.75% probability that the difference in mean weight of the samples is greater than fifteen pounds.

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 [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.

Subtraction of normal variables:

When we subtract two normal variables, the mean of the subtraction distribution is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

Scotts American Bulldog males have weights that are normally distributed with a mean of 85 pounds and a standard deviation of 14.6 pounds.

This means that [tex]\mu = 85, \sigma = 14.6[/tex]

Sample of 9 male bulldogs:

By the Central Limit Theorem:

[tex]n = 9, \mu_M = 85, s_M = \frac{14.6}{\sqrt{9}} = 4.8667[/tex]

Female American Bulldog have weights that are normally distributed with a mean of 69 pounds and a standard deviation of 8.8 pounds.

This means that [tex]\mu = 69, \sigma = 8.8[/tex]

Sample of 8 female bulldogs:

By the Central Limit Theorem:

[tex]n = 8, \mu_F = 69, s_F = \frac{8.8}{\sqrt{8}} = 3.1113[/tex]

Subtraction of Male by Female:

[tex]\mu_S = \mu_M - \mu_F = 85 - 69 = 16[/tex]

[tex]s_S = \sqrt{s_M^2+s_F^2} = \sqrt{4.8667^2+3.1113^2} = 5.7762[/tex]

What is the probability that the difference in mean weight of the samples (Male - Female) is greater than fifteen pounds?

This is 1 subtracted by the pvalue of Z when X = 15. So

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

With our distribution(subtraction)

[tex]Z = \frac{X - \mu_S}{s_S}[/tex]

[tex]Z = \frac{15-16}{5.7762}[/tex]

[tex]Z = -0.17[/tex]

[tex]Z = -0.17[/tex] has a pvalue of 0.4325

1 - 0.4325 = 0.5675

0.5675 = 56.75% probability that the difference in mean weight of the samples is greater than fifteen pounds.

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