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Alla [95]
3 years ago
12

A grocery store produce manager is told by a wholesaler that the apples in a large shipment have a mean weight of 6 ounces and a

standard deviation of 1.4 ounces. The manager is going to randomly select 49 apples. Suppose the manager is willing to risk a 10% chance of returning the shipment even if the wholesaler's claim is true. Let W be the mean weight of the sample below which the shipment will be returned. Find the value of W.
Mathematics
1 answer:
Andru [333]3 years ago
7 0

Answer:

W= 5.744

Step-by-step explanation:

given that a grocery store produce manager is told by a wholesaler that the apples in a large shipment have a mean weight of 6 ounces and a standard deviation of 1.4 ounces

Sample size n= 49

Margin of error = 0.10 (10% risk )

Let us assume X no of apples having mean weight of 6 oz is N(6,1.4)

Then sample mean will be normal with (6, 1.4/7) = (6,0.2)

(Because sample mean follows normal with std error as std dev /sqrt of sample size)

Now required probability <0.10

i.e.P(\bar X

Since x bar is normal we find z score for

P(Z

From std normal distribution table we find that z = 1.28

Corresponding X score =

W= 6-0.2*1.28\\W= 6-0.256 = 5.744

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Answer:

A is the only solution.

Step-by-step explanation:

A simple way to solve it is to plug in the x and y values

For A, we plug in 2 for x and 3 for y

3-3=5(2-2)

0=5(0)

0=0

Ordered pair A is a solution

For B, plug in 3 for x and 2 for y

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-1 does not equal 5

Therefore only A is a solution

5 0
3 years ago
One of my cats, Apollo, drives a cat car. He used 6/8 of a tank of gas to visit his friends for the weekend, and another 1 half
Arlecino [84]

Answer:

Total gas used = 3/2 of a tank of gas

Step-by-step explanation:

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7 0
3 years ago
6x + 9 &lt; 63<br> what is x?
Natalija [7]
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Subtract 9 from both sides.

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7 0
3 years ago
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In the last quarter of​ 2007, a group of 64 mutual funds had a mean return of 2.2 2.2​% with a standard deviation of 6.3 6.3​%.
Svet_ta [14]

Answer:

a) 0.625; b) 16.879; c) 8.442

Step-by-step explanation:

Since this is a normal distribution, we want the value of the mean, μ:  2.2; and the value of the standard deviation, σ:  6.3.

For the 40th percentile, we look in a z chart.  We want to find the value as close to 0.40 as we can get; this is 0.4013, and it corresponds to a z score of z = -0.25.

Our formula for a z score is z=\frac{X-\mu}{\sigma}.  Using our values, we have

-0.25 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(-0.25) = X-2.2

-1.575 = X-2.2

Add 2.2 to each side:

-1.575+2.2 = X-2.2+2.2

0.625 = X

For the 99th percentile, the value in the z chart closest to 0.99 is 0.9901, which corresponds to a z score of z = 2.33:

2.33 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(2.33) = X-2.2

14.679 = X-2.2

Add 2.2 to each side:

14.679+2.2 = X-2.2+2.2

16.879 = X

For the IQR, we find the values for the 75th percentile (Q3) and the 25th percentile (Q1).  The value in a z chart closest to 0.75 is 0.7486, which corresponds to a z score of z = 0.67:

0.67 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(0.67) = X-2.2

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Add 2.2 to each side:

4.221+2.2 = X-2.2+2.2

6.421 = X

The value in a z chart closest to 0.25 is 0.2514, which corresponds to a z score of z = -0.67:

-0.67 = (X-2.2)/6.3

Multiply both sides by 6.3:

6.3(-0.67) = X-2.2

-4.221 = X-2.2

Add 2.2 to each side:

-4.221+2.2 = X-2.2+2.2

-2.021 = X

This makes the interquartile range

6.421--2.021 = 8.442

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