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4 years ago

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The mean height of a Clydesdale horse is 72 inches with a standard deviation of 1.2 inches. What is the probability that a Clyde

sdale is greater than 75 inches tall?

1 answer:

Lubov Fominskaja [6]4 years ago

3 0

Answer:

0.0062

Step-by-step explanation:

Given that:

Mean (μ) = 72 inches, Standard deviation (σ) = 1.2 inches.

The z score is a measure in statistics is used to determine by how many standard deviation the raw score is above or below the mean. If the raw score is above the mean, the z score is positive and if the raw score is below the mean, the z score is negative.

The z score is given as:

$z=\frac{x-\mu}{\sigma}$

For Clydesdale is greater than 75 inches tall, x = 75 inches, the z score is:

$z=\frac{x-\mu}{\sigma}=\frac{75-72}{1.2} =2.5$

The probability that a Clydesdale is greater than 75 inches tall = P(X > 75) = P(Z > 2.5) = 1 - P(Z < 2.5) = 1 - 0.9938 = 0.0062 = 0.62%

The probability that a Clydesdale is greater than 75 inches tall is 0.62%

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Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

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3 years ago

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Step-by-step explanation:

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The fourth one seems like its the right one

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3 years ago