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Harlamova29_29 [7]
3 years ago
6

A survey found that​ women's heights are normally distributed with mean 63.4 in and standard deviation 2.2 in. A branch of the m

ilitary requires​ women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too​ tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest​ 1% and the tallest​ 2%, what are the new height​ requirements?
Mathematics
1 answer:
Ede4ka [16]3 years ago
8 0

Answer:

Step-by-step explanation:

Given that a survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.2 in.

Let X be the random variable representing women heights.

P(58<X<80)

Z for 58 = -2.454

Z for 80 = 7.545

P(-2.454<Z<7.545) = 0.4945+0.5=0.9945

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A soccer team sold raffle tickets to raise money for the upcoming season. They sold three different types of tickets: premium fo
lesya692 [45]

Answer:

45 premium tickets were sold

Step-by-step explanation:

p = premium

d = deluxe

r = regular

p+d+r = 273

6p+4d + 2r = 836

118+d = r

Replace r with 118+d

p+d+118+d = 273

p +2d = 273-118

p+2d = 155

6p+4d + 2(118+d) = 836

6p+4d + 236+2d = 836

6p +6d = 836-236

6p + 6d = 600

Divide by 6

p+d = 100

d = 100-p

Replace d in p +2d= 155

p +2(100-p) = 155

p+200-2p = 155

-p = 155-200

-p =-45

p =45

45 premium tickets were sold

3 0
3 years ago
Read 2 more answers
A class survey in a large class for first-year college students asked, “About how many minutes do you study on a typical weeknig
Anika [276]

Answer:  (110.22, 125.78)

Step-by-step explanation:

The confidence interval for the population mean is given by :-

\mu\ \pm z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}

Given : Sample size = 463

\mu=118\text{ minutes}

\sigma=65\text{ minutes}

Significance level : \alpha=1-0.99=0.01

Critical value : z_{\alpha/2}=z_{0.005}=\pm2.576

We assume that the population is normally distributed.

Now, the 90% confidence interval for the population mean will be :-

118\ \pm\ 2.576\times\dfrac{65}{\sqrt{463}} \\\\\approx118\pm7.78=(118-7.78\ ,\ 118+7.78)=(110.22,\ 125.78)

Hence, 99% confidence interval for the mean study time of all first-year students = (110.22, 125.78)

3 0
3 years ago
Converting Units Attachment Included
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Derman says, "I'm halfway through reading my book.if I read another 84 pages, I'll be two thirds of the way through my book." ho
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How do you find the area of a traizoid
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[(base + base)*high]/2

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