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dlinn [17]
3 years ago
6

Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the

probability that a randomly selected woman has a foot length less than 10.0 in.b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.
Mathematics
1 answer:
Makovka662 [10]3 years ago
5 0

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using 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.

In this problem, we have that:

\mu = 9.6, \sigma = 0.5.

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when X = 10.

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

Z = \frac{10 - 9.6}{0.5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

Z = \frac{8 - 9.6}{0.5}

Z = -3.2

Z = -3.2 has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1.

This probability is 1 subtracted by the pvalue of Z when X = 9.8. So:

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

Z = \frac{9.8 - 9.6}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

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d. sin A = 2\frac{\sqrt{13} }{13},  cos A = 3\frac{\sqrt{13} }{13},  tan A = \frac{2 }{3}

<h2>Step-by-step explanation:</h2>

The triangle for the question has been attached to this response.

As shown in the triangle;

AC = 36ft

BC = 24ft

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<em>Using Pythagoras' theorem;</em>

⇒ (AB)² = (AC)² + (BC)²

<em>Substitute the values of AC and BC</em>

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Ф = A

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<em>Substitute these values into equation (i) as follows;</em>

sin A = \frac{24}{12\sqrt{13} }

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<em>Substitute these values into equation (ii) as follows;</em>

cos A = \frac{36}{12\sqrt{13} }

cos A = \frac{3}{\sqrt{13}}

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<em>Substitute these values into equation (iii) as follows;</em>

tan A = \frac{24}{36}

tan A = \frac{2}{3}

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