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disa [49]
2 years ago
7

OFFERING 50 POINTS!!! The army reports that the distribution of waist sizes among female soldiers is approximately normal, with

a mean of 28.4 inches and a standard deviation of 1.2 inches. Part A: A female soldier whose waist is 26.1 inches is at what percentile? Explain your reasoning and justify your work mathematically. Part B: The army uniform supplier regularly stocks uniform pants between sizes 24 and 32. Anyone with a waist circumference outside that interval requires a customized order. Describe what this interval looks like if displayed visually. What percent of female soldiers requires custom uniform pants? Show your work and justify your reasoning mathematically.
Mathematics
1 answer:
hammer [34]2 years ago
4 0

Answer:

Step-by-step explanation:

1)

Percentile is related to the area under the standard normal curve to the LEFT of a certain data value (which in this case would be 26.1 inches).

On my Texas Instruments TI-83 Plus calculator, I found this area as follows:

normcdf(-100, 26.1, 28.4,1.2), where the range -100 to 26.1 represents the area (as a decimal fraction) to the left of 26.1 inches.  My result was 0.028, which corresponds to the 3rd percentile (0.028 rounds off to 0.03, which would be 3rd percentile).

2)  The mean waist size is 28.4 inches, represented by a vertical line through the standard normal curve lying between 24 and 32.  We use the same function on the calculator:  normcdf(24, 32, 28.4, 1.2).

The result is 0.9985.  Subtracting this from 1.0000, we get 0.001, or 0.1%, which is the percentage of female soldiers requiring custom uniforms.

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All numbers greater than or equal to -3.

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When planning a party you want to know whom to invite. Among the people you would like to invite are three touchy friends.You kn
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Jasmine only.

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No one.

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There is no possible way to invite all three friends.

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5 0
2 years ago
Suppose that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 poi
AnnZ [28]

Answer:

0.0266, 0.9997,0.7856

Step-by-step explanation:

Given that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 points. Also suppose that IQs of students from university B can be described by a normal model with mean 120 and standard deviation 11. Let x be the score by A students and Y the score of B.

A)P(X>135) = \\P(Z>0.625)\\=0.0266

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3 0
2 years ago
A customer at a store paid $64 for 3 large candles and 4 small candles. At the same store, a second customer paid $4 more than t
belka [17]

system of equations can be used to find the price in dollars of each large candle, x, and each small candle, y is  x+8y=68 and 3x+4y=64   .

<u>Step-by-step explanation:</u>

Here we have , A customer at a store paid $64 for 3 large candles and 4 small candles. At the same store, a second customer paid $4 more than the first customer for 1 large candle and 8 small candles. The price of each large candle is the same, and the price of each small candle is the same. We need to find Which system of equations can be used to find the price in dollars of each large candle, x, and each small candle, y . Let's find out:

Let the price in dollars of each large candle, x, and each small candle, y .So

A customer at a store paid $64 for 3 large candles and 4 small candles

Equation is  :

⇒ 3x+4y=64  .....(1)

At the same store, a second customer paid $4 more than the first customer for 1 large candle and 8 small candles.

Equation is  :

⇒ x+8y=68  .......(2)

3(2)-(1) i.e.

⇒ 3(x+8y)-(3x+4y)=68(3)-64

⇒ 20y=140

⇒ y=7

So , x+8y=68  

⇒  x+8(7)=68

⇒  x=12

Therefore , system of equations can be used to find the price in dollars of each large candle, x, and each small candle, y is  x+8y=68 and 3x+4y=64   .

7 0
2 years ago
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