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Inessa [10]
3 years ago
15

Professors at a local university earn an average salary of $80,000 with a standard deviation of $6,000. The salary distribution

cannot be regarded as bell-shaped. What can be said about the percentage of salaries that are less than $68,000 or more than $92,000?
a. It is at least 75 percent.
b. It is at least 55 percent.
c. It is at least 25 percent.
d. It is at most 25 percent.
Mathematics
1 answer:
Sladkaya [172]3 years ago
7 0

Answer:

d. It is at most 25 percent.

Step-by-step explanation:

According to Chebyshev's inequality, for a given number of standard deviations, k, no more than 1/k² can be more than k standard deviations from the mean. In this situation the amount of standard deviations from the mean of the upper and lower bound of salaries are:

U=\frac{\$92,000-\$80,000}{\$6,000}=2\\L = \frac{\$80,000-\$68,000}{\$6,000}= 2

For k = 2, applying Chebyshev's inequality:

P( \$68,000 \leq X \leq \$92,000)= \frac{1}{2^2} = 0.25

Therefore, at most 25% of the salaries are less than $68,000 or more than $92,000.

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Answer:

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Step-by-step explanation:

Given :

The average summer temperature in Anchorage is 69°F.

The daily temperature is normally distributed with a standard deviation of 7°F .

To Find:What percentage of the time would the temperature be between 55°F and 76°F?

Solution:

Mean = \mu = 69

Standard deviation = \sigma = 7

Formula : z=\frac{x-\mu}{\sigma}

Now At x = 55

z=\frac{55-69}{7}

z=-2

At x = 76

z=\frac{76-69}{7}

z=1

Now to find P(55<z<76)

P(2<z<-1)=P(z<2)-P(z>-1)

Using z table :

P(2<z<-1)=P(z<2)-P(z>-1)=0.9772-0.1587=0.8185

Now percentage of the time would the temperature be between 55°F and 76°F = 0.8185 \times 100 = 81.85\%

Hence If the daily temperature is normally distributed with a standard deviation of 7°F, 81.85% of the time would the temperature be between 55°F and 76°F.

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