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RUDIKE [14]
3 years ago
8

Women's heights are normally distributed with mean 63.5 in and standard deviation of 2.5 in. A social organization for tall peop

le has a requirement that women must be at least 70 in tall. What percentage of women meet that requirement?
Mathematics
1 answer:
Vanyuwa [196]3 years ago
8 0
First, ask yourself how far way is 70 from the mean of 63.5? It is equal to 70 - 63.5 = 6.5 inches. 

Next, ask yourself how many standard deviations is this? Well, every standard deviation equals 2.5 inches, so there are 6.5 / 2.5 = 2.6 standard deviations. So, 70 is exactly 2.6 standard deviations to the right of the mean. 

Finally, you can use the Standard Normal to find the answer P(X>70) since this is equivalent to: 

<span>P(z > 2.6) = 0.0054 or 0.54% of women will be taller than 70 inches. 

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Answer: The dimensions of the frame:.

W= 2√10 or about 6.3245 inches

L= 4√10 or about 12.6491 inches

Step-by-step explanation:

This would be so much easier if the area of the box was 72. Then the dimensions of the frame would be 6 × 12. So just a little bit larger and more complicated.

Here the equation is w(2w)=80

2w^2 = 80 (2w is easier to work with than L/2).

w^2=40 (^2 means squared)

√w^2 = √40 square root of both sides: Factor 40. Take out√4

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The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co
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Answer:

95% Confidence interval for the variance:

3.6511\leq \sigma^2\leq 34.5972

95% Confidence interval for the standard deviation:

1.9108\leq \sigma \leq 5.8819

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

s^2=2.89^2=8.3521

The confidence interval for the variance is:

\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128

Then, the confidence interval can be calculated as:

\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819

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