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As reported in Runner’s World magazine, the times of the finishers in the New York City...

  1. As reported in Runner’s World magazine, the times of the finishers in the New York City 10-km run are Normally distributed with mean 61 minutes and standard deviation 9 minutes.
    a. What percentage of runners finish with a time less than 75 minutes?
    b. What percentage of runners finish with a time between 50 and 70 minutes?

    c. What percentage of runners take longer than 90 minutes to finish?

  2. The annual salaries or employees in a large company are approximately Normally distributed with a mean value of $50 thousand and a standard deviation of $20 thousand. a. What percentage of employees earn less than $40 thousand? b. What percentage of employees earn between $45 and $65 thousand? c. What percentage of employees earn more than $70,000? d. If an employee is in the bottom 2% of all salaries, what must his/her salary be? e. What would an employee have to earn to be in the top 1% of all salaries?

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Answer #1

Solution :

Given that ,

mean = = 61

standard deviation = = 9

a). What percentage of runners finish with a time less than 75 minutes?

Ans: P(X<75) = P[(X- ) / < (75-61) / 9]

= P(z <1.55 )

Using z table

= 0.94009

=94.01%

b. What percentage of runners finish with a time between 50 and 70 minutes?

Ans:

P(50< x < 70) = P[(50- 61) / 9< (x - ) / < (70-61) / 9)]

= P(-1.22 < Z <1 )

= P(Z <1 ) - P(Z < -1.22)

Using z table   

= 0.8413-0.1112

=0.7301

=73.01%

c. What percentage of runners take longer than 90 minutes to finish?

Ans:P(X>90)=1-P(X<90)

P(X>90) =1- P[(X- ) / < (90-61) / 9]

= 1-P(z <3.22 )

Using z table

=1- 0.9994

=0.0006

=0.06%

#Solution b

mean = = 50

standard deviation = = 20

a. What percentage of employees earn less than $40 thousand?

Ans: P(X<40) = P[(X- ) / < (40-50) / 20]

= P(z <-0.5 )

Using z table

= 0.3085

=30.85%

b. What percentage of employees earn between $45 and $65 thousand?

Ans:P(45< x < 65) = P[ (45-50) / 20< (x - ) / < (65-50) / 20)]

= P(-0.25 < Z <0.75 )

= P(Z <0.75 ) - P(Z < -0.25)

Using z table   

= 0.7734-0.4013

=0.3721

=37.21%

As per HomeworkLib policy we solve only 1st four sub-part

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