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aalyn [17]
2 years ago
15

18. A normal population has a mean of 80.0 and a standard deviation of 14.0. a. Compute the probability of a value between 75.0

and 90.0. b. Compute the probability of a value 75.0 or less. c. Compute the probability of a value between 55.0 and 70.0. Lind, Douglas. Basic Statistics for Business and Economics (p. 213). McGraw-Hill Education. Kindle Edition.
Mathematics
1 answer:
mixer [17]2 years ago
6 0

Answer:

a) 40.17% probability of a value between 75.0 and 90.0.

b) 35.94% probability of a value 75.0 or less.

c) 20.22% probability of a value between 55.0 and 70.0.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 80, \sigma = 14

a. Compute the probability of a value between 75.0 and 90.0.

This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 75.

X = 90

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

Z = \frac{90 - 80}{14}

Z = 0.71

Z = 0.71 has a pvalue of 0.7611

X = 75

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

Z = \frac{75 - 80}{14}

Z = -0.36

Z = -0.36 has a pvalue of 0.3594

0.7611 - 0.3594 = 0.4017

40.17% probability of a value between 75.0 and 90.0.

b. Compute the probability of a value 75.0 or less.

This is the pvalue of Z when X = 75. So

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

Z = \frac{75 - 80}{14}

Z = -0.36

Z = -0.36 has a pvalue of 0.3594

35.94% probability of a value 75.0 or less.

c. Compute the probability of a value between 55.0 and 70.0.

This is the pvalue of Z when X = 70 subtracted by the pvalue of Z when X = 55.

X = 70

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

Z = \frac{70 - 80}{14}

Z = -0.71

Z = -0.71 has a pvalue of 0.2389

X = 55

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

Z = \frac{55 - 80}{14}

Z = -1.79

Z = -1.791 has a pvalue of 0.0367

0.2389 - 0.0367 = 0.2022

20.22% probability of a value between 55.0 and 70.0.

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Answer:

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Step-by-step explanation:

we know that

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Step 2

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Answer:

<u>The distance of the library in the map is 0.67 inches (Rounding to two decimal places)</u>

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1. Let's review all the information given for solving this question:

Scale factor for the map of the town = 0.5 inches: 3 miles

Distance from Niko's house to his school = 3 miles

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2. Let's find the distance of the library in the map

Distance of the library in the map = Distance of the library * Scale factor

Distance of the library in the map = 4 miles * 0.5 inches:3 miles

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19 small mowers and 11 large mowers are sold

<em><u>Solution:</u></em>

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Let "b" be the number of large mowers sold

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Cost of each large mower = $ 329.99

<em><u>30 total mowers were sold</u></em>

Therefore,

a + b = 30

a = 30 - b ------------- eqn 1

<em><u>The total sales for a given year was $8379.70</u></em>

<em><u>Thus we frame a equation as:</u></em>

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a \times 249.99 + b \times 329.99 = 8379.70

249.99a + 329.99b = 8379.70 ---------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Substitute eqn 1 in eqn 2</u></em>

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80b = 8379.70 - 7499.7

80b = 880

Divide both sides by 80

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Answer:

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c. d = 2, u1 = 9

Step-by-step explanation:

a. The given parameters are;

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The amount of drug increase received by Kiri each day = d

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The amount of drug she received on the eleventh day = 29 mg

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aₙ = a₁ + (n - 1)·d

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a₁ = u1

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a₇ = u1 + (7 - 1)·d = u1 + 6·d = 21

The equation for the amount of drugs she receives in terms of u1 and d on the seventh day is given as follows;

21 = u1 + 6·d

b. For the eleventh day, the amount of drugs she receives, which is 29 milligrams, is given as follows;

a₁₁ = u1 + (11 - 1)·d = u1 + 10·d = 29

Therefore, the equation for the amount of drugs she receives in terms of u1 and d on the  eleventh day is given as follows;

29 = u1 + 10·d

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29 = u1 + 10·d..............(2)

Subtracting equation (1) from equation (2) gives;

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8 = 4·d

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21 = u1 + 12

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d = 2, u1 = 9.

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BigorU [14]

Answer:

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Then the expected value of <em>X</em> is:

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The variance of the random variable <em>X</em> is:

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(b)

The random variable <em>Z</em> is the standardized form of the random variable <em>X</em>.

It is defined as:Z=\frac{X-\mu}{\sigma}

Compute the expected value of <em>Z</em> as follows:

E(Z)=E[\frac{X-\mu}{\sigma}]\\=\frac{E(X)-\mu}{\sigma}\\=\frac{61-61}{9}\\=0

The mean of <em>Z</em> is 0.

Compute the variance of <em>Z</em> as follows:

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The variance of <em>Z</em> is 1.

So the standard deviation is 1.

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