Question

1. Mr. Mcee hosts a campus contest to see who can listen to the 80’s song by Barnes & Barnes titled “Fish Heads” the most number of times in a row without collapsing in agony. One hundred students are selected at random to participate. Upon reviewing the data, Dr. Mcee discovers that that the mean number of song plays before a student would collapse in agony is 85 song plays, and the standard deviation is 9 song plays. Please answer the following questions about the UP student body:

1. a) How many times could a randomly-selected UP student listen to this terrible song without collapsing, to fall within the middle 95% of all students?

2. b) How many song plays separate the top 0.5% of the students—the truly resilient who can endure great pain and suffering—from everyone else?

3. c) What if the sample size was 600, and the standard deviation was 25? What range of song plays would capture the middle 99% of the UP population?

4. d) What would your answer for part c be, if McRee only had 20 students participate in the contest?

Answer 1

μ = 85

σ = 9

n=100

A)

95% confidence z= +/- 1.96 [standard normal distribution table]

Z1=X1-μ/(σ / √n)

-1.96= X1-85 /(9/√100)

X1=83.236 =83

Z2=X2-μ/(σ / √n)

1.96= X2-85 /(9/√100)

X2= 86.764 =87

Number of times could a randomly-selected UP student listen to this terrible song without collapsing is 83 to 87

B) from standard normal distribution table

For top 0.5% of the students z= 2.575

Z=X-μ/(σ / √n)

2.575 = x- 85/(9/√100)

X= 87.3175 =87

87 songs sepa2 top 0.5% of the students

C)

n=600

σ = 25

99% confidence z= +/- 2.575 [standard normal distribution table]

Z1=X1-μ/(σ / √n)

-2.575= X1-85 /(25/√600)

X1=82.37

=82

Z2=X2-μ/(σ / √n)

2.575= X2-85 /(25/√600)

X2= 87.63 =88

Range (82,88)

D)

n=20

σ = 25

99% confidence z= +/- 2.575 [standard normal distribution table]

Z1=X1-μ/(σ / √n)

-2.575= X1-85 /(25/√20)

X1=70.61

=71

Z2=X2-μ/(σ / √n)

2.575= X2-85 /(25/√600)

X2= 99.39

=99

Range (71,99)