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3 years ago

15

5 times the difference between 19 and 13

1 answer:

BaLLatris [955]3 years ago

6 0

5 times 19-13 would equal 30 as 19-13 is 6 times 5 =30

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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

Makovka662 [10]

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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