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2 years ago

8

William earned $3,200 per month as a teacher for the ten months from September to June. Then he took a job as a barista at a loc

al café, where he earned $2,000 per month during July and August. What was his average monthly pay for the 12 months?

1 answer:

Kitty [74]2 years ago

7 0

Answer:

$3,000

Step-by-step explanation:

To find his average monthly pay, you have to find the total amount that Willian received in a year by multiplying his monthly salary as a teacher for 10 months and his salary as a barista for 2 months and adding up these numbers. Then, you have to divide the total amount he earned by 12 that is the number of months in a year:

($3,200*10)+($2,000*2)= $32,000+4,000=36,000

$36,000/12=$3,000

According to this, his average monthly pay for the 12 months was $3,000.

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A family purchases a light sphere to be used out on their patio. The diameter of the sphere is 20 in.

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Step-by-step explanation:

Formula

$Volume\ of\ a\ sphere = \frac{4}{3}\pi\ r^{3}$

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3 years ago

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United Sta

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a) $5,656.85

b) Bell-shaped(normally distributed).

c) 36.32% probability of selecting a sample with a mean of at least $112,000.

d) 96.16% probability of selecting a sample with a mean of more than $100,000.

e) 59.84% probability of selecting a sample with a mean of more than $100,000 but less than $112,000.

Step-by-step explanation:

To solve this question, it is important 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 $\sigma$ , a large sample size, of size 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:

$\mu = 110000, \sigma = 40000$

a. If we select a random sample of 50 households, what is the standard error of the mean?

This is the standard deviation of the sample, that is, s, when $n = 50$ .

So

$s = \frac{\sigma}{\sqrt{n}} = \frac{40000}{\sqrt{50}} = 5656.85$

b. What is the expected shape of the distribution of the sample mean?

By the Central Limit Theorem, bell-shaped(normally distributed).

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

This is 1 subtracted by the pvalue of Z when X = 112000. So

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

By the Central Limit Theorem

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

$Z = \frac{112000 - 110000}{5656.85}$

$Z = 0.35$

$Z = 0.35$ has a pvalue of 0.6368

So 1-0.6368 = 0.3632 = 36.32% probability of selecting a sample with a mean of at least $112,000.

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

This is 1 subtracted by the pvalue of Z when X = 112000. So

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

$Z = \frac{100000 - 110000}{5656.85}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

So 1-0.0384 = 0.9616 = 96.16% probability of selecting a sample with a mean of more than $100,000.

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

This is the pvalue of Z when X = 112000 subtractex by the pvalue of Z when X = 100000.

So

X = 112000

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

$Z = \frac{112000 - 110000}{5656.85}$

$Z = 0.35$

$Z = 0.35$ has a pvalue of 0.6368

X = 100000

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

$Z = \frac{100000 - 110000}{5656.85}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

So 0.6368 - 0.0384 = 0.5984 = 59.84% probability of selecting a sample with a mean of more than $100,000 but less than $112,000.

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3 years ago