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3 years ago

The graph shows a distribution of data.

Mathematics

2 answers:

polet [3.4K]3 years ago

6 0

Answer:

the answer is 0.0625 (a)

Step-by-step explanation:

i took the test on edginuity

ss7ja [257]3 years ago

5 0

Answer:

.0625 or A

Step-by-step explanation:

You have to work out the mean. (The simple average of those numbers) Then for each number subtract the Mean and square the result. Then work out the average of those squared differences.

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3 years ago

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3 years ago

According to the U.S. Bureau of Labor Statistics, 20% of all people 16 years of age or older do volunteer work. In this age grou

Murljashka [212]

Answer:

1. P(X≥35) = 0.0183

2. P(X≤21) = 0.0183

3. P(0.18<p<0.25) = 0.7915

Step-by-step explanation:

We have the proportion for women: pw=0.22, and the proportion for men: pm=0.19.

1. We have a sample of 140 woman and we have to calculate the probability of getting 35 or more who do volunteer work.

This is equivalent to a proportion of

$p=X/n=35/140=0.25$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22*0.78}{300}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.0239$

We calculate the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.25-0.22}{0.0239}=\dfrac{0.03}{0.0239}=0.8198$

Then, the probability of having 35 women or more who do volunteer work in this sample of 140 women is:

$P(X>35)=P(p>0.25)=P(z>2.0906)=0.0183$

2. We have to calculate the probability of having 21 or fewer women in the group who do volunteer work.

The proportion is now:

$p=X/n=21/140=0.15$

We can calculate then the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.15-0.2}{0.0239}=\dfrac{-0.05}{0.0239}=-2.0906$

Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:

$P(X$

3. For the sample with men and women, we use the proportion for both, which is π=0.2.

The sample size is n=300.

Then, the standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{300}}\\\\\\ \sigma_p=\sqrt{0.0005}=0.0231$

We can calculate the z-scores for p1=0.18 and p2=0.25:

$z_1=\dfrac{p_1-\pi}{\sigma_p}=\dfrac{0.18-0.2}{0.0231}=\dfrac{-0.02}{0.0231}=-0.8660\\\\\\z_2=\dfrac{p_2-\pi}{\sigma_p}=\dfrac{0.25-0.2}{0.0231}=\dfrac{0.05}{0.0231}=2.1651$

We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:

$P=P(0.18$

5 0

3 years ago