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Alexxx [7]
3 years ago
14

Completion time (from start to finish) of a building remodeling project is normally distributed with a mean of 200 work-days and

a standard deviation of 10 work-days. To be 99% sure that we will not be late in completing the project, we should request a completion time of _______ work-days.
Mathematics
2 answers:
-BARSIC- [3]3 years ago
5 0
We should request a completion time of

223 work-days.
stiks02 [169]3 years ago
5 0

Answer:

233 days.

Step-by-step explanation:

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 = 200, \sigma = 10

To be 99% sure that we will not be late in completing the project, we should request a completion time of ...

This is the value of X when Z has a pvalue of 0.99. So X when Z = 2.325.

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

2.325 = \frac{X - 200}{10}

X - 200 = 10*2.325

X = 232.5

So the correct answer is 233 days.

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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:

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

P=P(0.18

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