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Komok [63]
3 years ago
7

The heights of women aged 20 to 29 are approximately normal with mean 64 inches and standard deviation 2.7 inches. Men the same

age have mean height 69.3 inches with standard deviation 2.8 inches. What are the z-scores for a woman 6 feet tall and a man 6 feet tall
Mathematics
1 answer:
ohaa [14]3 years ago
3 0

Answer:

z=(x-mu)/o

Step-by-step explanation:

X - Heights of women aged 20 to 29: X is N(64, 2.7)

Y - Heights of men aged 20 to 29: X is N(69.3, 2.8)

a) X=6'=72"

b) Y = 72"

c) 72' women can be taken as very tall but men moderately taller than average

d) Y<5'5" =65"

e) X>73"

Z>

Almost 0% can be taken

f) IQR =Q3-Q1

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141.4 would be your answer
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Over Thanksgiving Break Joshua drove from Connecticut to Ohio which is 422 miles. This trip took Joshua and his family six hours
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70 miles per hour

Step-by-step explanation:

422 divided by 6

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3 years ago
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The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml
irina [24]

Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, we need 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, the sample means with size n of 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 = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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

Z = \frac{120.1 - 119}{1.63}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

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

Z = \frac{117.9 - 119}{1.63}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

8 0
3 years ago
the length of a rectangular patio is 8 feet less than twice its width. the area of the patio is 280 square feet. find the dimens
jolli1 [7]

Answer:

The length of the rectangle 'l' = 20

The width of the rectangle 'w' = 14

Step-by-step explanation:

<u>Explanation</u>:-

Let 'x' be the width

Given data the length of a rectangular patio is 8 feet less than twice its width

2x-8 = length

The area of rectangle = length X width

Given area of rectangle = 280 square feet

x(2x-8) = 280

2(x)(x-4) =280

x(x-4) =140

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(x+10)(x-14) =0

x = -10 and x = 14

we can choose only x =14

The width of the rectangle 14

The length of the rectangle 2x-8 = 2(14)-8 = 28 -8 =20

The length of the rectangle 'l' = 20

The width of the rectangle 'w' = 14

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2/22.
i think
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