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denis23 [38]
3 years ago
8

The lengths of a lawn mower part are approximately normally distributed with a given mean = 4 in. and standard deviation = 0.2 i

n. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.? 34% 68% 95% 99.7%
Mathematics
2 answers:
vova2212 [387]3 years ago
8 0

Answer: EDGE 2020 Quiz

Step-by-step explanation:

1. The average miles per gallon of a particular automobile model are approximately normally distributed with a given mean Mu = 43.8 miles per gallon and standard deviation Sigma = 5.1 miles per gallon. What percentage of the automobiles have an average miles per gallon between 38.7 miles per gallon and 48.9 miles per gallon?

A.) 68%

2. What is the mean of the normal distribution shown below?

C.) 4

3. The annual salaries of all employees at a financial company are normally distributed with a mean of $34,000 and a standard deviation of $4,000. What is the z-score of a company employee who makes an annual salary of $54,000?

D.) 5

4. The lengths of a lawn mower part are approximately normally distributed with a given mean Mu = 4 in. and standard deviation Sigma = 0.2 in. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.?

B.) 68%

5. The mean of the temperatures in the chart is 24° with a standard deviation of 4°. Which temperature is within one standard deviation of the mean?

C.) 22°

6. Which statement must be true?

D.) Each distribution has the same mean and a different standard deviation.

7. The heights of a certain type of tree are approximately normally distributed with a mean height Mu = 5 ft and a standard deviation Sigma = 0.4 ft. Which statement must be true?

D.) A tree with a height of 6.2 ft is 3 standard deviations above the mean.

8. What is the difference of the means of the distributions?

A.) 15

9. )The weights of boxes of candies produced in a factory are normally distributed with a mean weight of 16 oz and a standard deviation of 1 oz. What is the weight of a box of candies with a z-score of 2?

B.) 18 oz

10. What is the mean of the normal distribution shown below?

B.) 0

Charra [1.4K]3 years ago
5 0

Answer:

68 %

Step-by-step explanation:

Let's call X to the random variable ''lengths of a lawn mower part''.

X ~ N(given mean;standard deviation)

X ~ N(4 in;0.2 in)

To find the percentage,first we are going to turn this random variable X into a random variable Z. Z will be a N(0;1)

We do this by subtracting the given mean to the original variable and dividing by it deviation.

P (3.8 in < X < 4.2 in) =

P [(3.8 in - 4 in) / (0.2 in)] < [(X - 4 in)] / (0.2 in) < [(4.2 in - 4 in) / (0.2 in)]

P ( -1 < Z < 1 ) = P (Z < 1) - P (Z< -1)

Where we can find  P (Z < 1) and  P (Z< -1) in any table of a N(0;1)

P (Z < 1) - P (Z< -1) = 0.8413 - 0.1587 = 0.6826 = 68%

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Write <br> 108<br> 13<br> as a mixed number.
gayaneshka [121]

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8 4/13 Is the Correct Answer!

Step-by-step explanation:

Numerator. This is the number above the fraction line. For 108/13, the numerator is 108.

Denominator. This is the number below the fraction line. For 108/13, the denominator is 13.

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Now let's go through the steps needed to convert 108/13 to a mixed number.

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Now that we have our whole number for the mixed fraction, we need to find our new numerator for the fraction part of the mixed number.

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Step 4: Simplifying our fraction

In this case, our fraction (4/13) can be simplified down further. In order to do that, we need to calculate the GCF (greatest common factor) of those two numbers. You can use our handy GCF calculator to work this out yourself if you want to. We already did that, and the GCF of 4 and 13 is 1.

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3
-
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x
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+
10
=
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Rewrite
x
2
3
x
2
3
as
(
x
1
3
)
2
(
x
1
3
)
2
.
(
x
1
3
)
2
−
7
x
1
3
+
10
=
0
(
x
1
3
)
2
-
7
x
1
3
+
10
=
0
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u
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