Mathematics, 11.06.2020 20:57 belle5828

Suppose that a population of people has an average weight of 160 lbs, and standard deviation of 50 lbs, and that weight is normally distributed. A researcher samples 100 people, and measures their weight. Find the probability that the researcher observes an average weight of the 100 people to be between 150 and 170. [Round your answer to four decimal places]

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Answer from: cammiecam49

0.9544 = 95.44% probability that the researcher observes an average weight of the 100 people to be between 150 and 170.

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 normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 160, \sigma = 50, n = 100, s = \frac{50}{\sqrt{100}} = 5$

Find the probability that the researcher observes an average weight of the 100 people to be between 150 and 170.

This is the pvalue of Z when X = 170 subtracted by the pvalue of Z when X = 150. So

X = 170

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

By the Central Limit Theorem

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

$Z = \frac{170 - 160}{5}$

Z = 2

Z = 2 has a pvalue of 0.9772

X = 150

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

$Z = \frac{150 - 160}{5}$

Z = -2

Z = -2 has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the researcher observes an average weight of the 100 people to be between 150 and 170.

Answer from: Quest

man its nopq i just guessed and got it right so uhh

g00d lucc

step-by-step explanation:

Answer from: Quest

10 you would take 50/ .2

step-by-step explanation:

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