Mathematics, 17.03.2020 02:05 enthusiastfranklin

A leading magazine (like Barron's) reported at one time that the average number of weeks an individual is unemployed is 29 weeks. Assume that the length of unemployment is normally distributed with population mean of 29 weeks and the population standard deviation of 9 weeks. Suppose you would like to select a random sample of 38 unemployed individuals for a follow-up study.

a. Find the probability that a single randomly selected value is greater than 27.6.
b. Find the probability that a sample of size n=80 is randomly selected with a mean greater than 27.6.

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Answer from: jurnee77

a) 83.15% probability that a single randomly selected value is greater than 27.6.

b) 91.92% probability that a sample of size n=80 is randomly selected with a mean greater than 27.6.

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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 29, \sigma = 9, n = 38, s = \frac{9}{\sqrt{38}} = 1.46$

a. Find the probability that a single randomly selected value is greater than 27.6.

This is 1 subtracted by the pvalue of Z when X = 27.6. So

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

By the Central Limit Theorem

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

$Z = \frac{27.6 - 29}{1.46}$

Z = -0.96

Z = -0.96 has a pvalue of 0.1685

1 - 0.1685 = 0.8315

83.15% probability that a single randomly selected value is greater than 27.6.

b. Find the probability that a sample of size n=80 is randomly selected with a mean greater than 27.6.

Now we have $n = 80, s = \frac{9}{\sqrt{80}} = 1$

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

$Z = \frac{27.6 - 29}{1}$

Z = -1.4

Z = -1.4 has a pvalue of 0.0808.

1 - 0.0808 = 0.9192

91.92% probability that a sample of size n=80 is randomly selected with a mean greater than 27.6.

Answer from: Quest

the best buy, is the 4- kilogram bag of lollipops for $18.40.

step-by-step explanation:

i found my answer by dividing the cost of each bag by their respective weights.

14.88 divided by 2 = 7.44

17.29 divided by 3 = 5.77

18.4 divided by 4 = 4.6

24.4 divided by 5 = 4.88

by this, i found that the 4 kilo bag is the best buy.

Answer from: Quest

6hope this ! ! ’

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