STAT 200 Week 4 Homework Problems
6.1.2
1.) The commuter trains on the Red Line for the Regional Transit Authority (RTA) in Cleveland,
OH, have a waiting time during peak rush hour periods of eight minutes ("2012 annual report,"
2012).
a.) State the random variable.
x=waiting time during peak hours
b.) Find the height of this uniform distribution.
1/(8-0)=0.125
c.) Find the probability of waiting between four and five minutes.
P(4<x<5)=(5-4)*0.125=0.125
d.) Find the probability of waiting between three and eight minutes.
P(3<x<8)=(8-3)*0.125=0.625
e.) Find the probability of waiting five minutes exactly.
P(x=5)=0*0.125=0.000
6.3.2
Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal
distribution with mean of and standard deviation .
a.) The area to the left of z is 15%.
Computation in google sheets =NORMINV(0.15,0,1)= -1.036
b.) The area to the right of z is 65%.
Computation in google sheets =NORMINV (0.35,0,1)= -0.385
c.) The area to the left of z is 10%.
Computation in google sheets =NORMINV(0.1,0,1)= -1.282
d.) The area to the right of z is 5%.
Computation in google sheets =NORMINV(0.95,0,1)= 1.645
e.) The area between and z is 95%. (Hint draw a picture and figure out the area to the left
of the .)
Computation in google sheets =NORMSINV(0.95)= -1.645;1.645
f.) The area between and z is 99%.
Computation in google sheets =NORMSINV(0.99)=-2.326;2.326
6.3.4
According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with
a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Assume that blood pressure is
normally distributed.
a.) State the random variable.
x=blood pressure for people in China.
b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.
calculator normalcdf(135,1e99,128,23) = 0.3804
, c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.
calculator normalcdf(-1e99,141,128,23) = 0.714
d.) Find the probability that a person in China has blood pressure between 120 and 125 mmHg.
P(120<x<125)
normalcdf(120,125,128,23) = 0.0841
e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?
From question b., it is unusual because the probability of blood pressure of 135 mmHg or
more is less than 5%.
f.) What blood pressure do 90% of all people in China have less than?
From google sheets =norminv(0.9,128,23) = 157.476
6.3.8
A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance
life expectancy," 2013). Assume the life of a dishwasher is normally distributed.
a.) State the random variable.
x= life expectancy of a dishwasher
b.) Find the probability that a dishwasher will last more than 15 years.
calculator normalcdf(15,1e99,12,1.25) = 0.008
c.) Find the probability that a dishwasher will last less than 6 years.
calculator normalcdf(-1e99,6,12,1.25) = 0
d.) Find the probability that a dishwasher will last between 8 and 10 years.
calculator normalcdf(8,10,12,1.25) = 0.054
e.) If you found a dishwasher that lasted less than 6 years, would you think that you have a
problem with the manufacturing process? Why or why not?
Yes, because from question c. the probability of having a dishwasher with a life expectancy
of 6 years or less is 0.
f.) A manufacturer of dishwashers only wants to replace free of charge 5% of all dishwashers.
How long should the manufacturer make the warranty period?
From google sheets =norminv(0.05,12,1.25) = 9.944
6.3.10
The mean yearly rainfall in Sydney, Australia, is about 137 mm and the standard deviation is about 69
mm ("Annual maximums of," 2013). Assume rainfall is normally distributed.
a.) State the random variable.
x=rainfall quantity
b.) Find the probability that the yearly rainfall is less than 100 mm.
calculator normalcdf(-1e99,100,137,69) = 0.296
c.) Find the probability that the yearly rainfall is more than 240 mm.
calculator normalcdf(240,1e99,137,69) = 0.068
d.) Find the probability that the yearly rainfall is between 140 and 250 mm.
calculator normalcdf(140,250,137,69) = 0.432
e.) If a year has a rainfall less than 100mm, does that mean it is an unusually dry year? Why or
why not?
Yes, because from question b. the probability is less than 5%
f.) What rainfall amount are 90% of all yearly rainfalls more than?
6.1.2
1.) The commuter trains on the Red Line for the Regional Transit Authority (RTA) in Cleveland,
OH, have a waiting time during peak rush hour periods of eight minutes ("2012 annual report,"
2012).
a.) State the random variable.
x=waiting time during peak hours
b.) Find the height of this uniform distribution.
1/(8-0)=0.125
c.) Find the probability of waiting between four and five minutes.
P(4<x<5)=(5-4)*0.125=0.125
d.) Find the probability of waiting between three and eight minutes.
P(3<x<8)=(8-3)*0.125=0.625
e.) Find the probability of waiting five minutes exactly.
P(x=5)=0*0.125=0.000
6.3.2
Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal
distribution with mean of and standard deviation .
a.) The area to the left of z is 15%.
Computation in google sheets =NORMINV(0.15,0,1)= -1.036
b.) The area to the right of z is 65%.
Computation in google sheets =NORMINV (0.35,0,1)= -0.385
c.) The area to the left of z is 10%.
Computation in google sheets =NORMINV(0.1,0,1)= -1.282
d.) The area to the right of z is 5%.
Computation in google sheets =NORMINV(0.95,0,1)= 1.645
e.) The area between and z is 95%. (Hint draw a picture and figure out the area to the left
of the .)
Computation in google sheets =NORMSINV(0.95)= -1.645;1.645
f.) The area between and z is 99%.
Computation in google sheets =NORMSINV(0.99)=-2.326;2.326
6.3.4
According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with
a standard deviation of 23 mmHg (Kuulasmaa, Hense & Tolonen, 1998). Assume that blood pressure is
normally distributed.
a.) State the random variable.
x=blood pressure for people in China.
b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.
calculator normalcdf(135,1e99,128,23) = 0.3804
, c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.
calculator normalcdf(-1e99,141,128,23) = 0.714
d.) Find the probability that a person in China has blood pressure between 120 and 125 mmHg.
P(120<x<125)
normalcdf(120,125,128,23) = 0.0841
e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?
From question b., it is unusual because the probability of blood pressure of 135 mmHg or
more is less than 5%.
f.) What blood pressure do 90% of all people in China have less than?
From google sheets =norminv(0.9,128,23) = 157.476
6.3.8
A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance
life expectancy," 2013). Assume the life of a dishwasher is normally distributed.
a.) State the random variable.
x= life expectancy of a dishwasher
b.) Find the probability that a dishwasher will last more than 15 years.
calculator normalcdf(15,1e99,12,1.25) = 0.008
c.) Find the probability that a dishwasher will last less than 6 years.
calculator normalcdf(-1e99,6,12,1.25) = 0
d.) Find the probability that a dishwasher will last between 8 and 10 years.
calculator normalcdf(8,10,12,1.25) = 0.054
e.) If you found a dishwasher that lasted less than 6 years, would you think that you have a
problem with the manufacturing process? Why or why not?
Yes, because from question c. the probability of having a dishwasher with a life expectancy
of 6 years or less is 0.
f.) A manufacturer of dishwashers only wants to replace free of charge 5% of all dishwashers.
How long should the manufacturer make the warranty period?
From google sheets =norminv(0.05,12,1.25) = 9.944
6.3.10
The mean yearly rainfall in Sydney, Australia, is about 137 mm and the standard deviation is about 69
mm ("Annual maximums of," 2013). Assume rainfall is normally distributed.
a.) State the random variable.
x=rainfall quantity
b.) Find the probability that the yearly rainfall is less than 100 mm.
calculator normalcdf(-1e99,100,137,69) = 0.296
c.) Find the probability that the yearly rainfall is more than 240 mm.
calculator normalcdf(240,1e99,137,69) = 0.068
d.) Find the probability that the yearly rainfall is between 140 and 250 mm.
calculator normalcdf(140,250,137,69) = 0.432
e.) If a year has a rainfall less than 100mm, does that mean it is an unusually dry year? Why or
why not?
Yes, because from question b. the probability is less than 5%
f.) What rainfall amount are 90% of all yearly rainfalls more than?
