- 31st Aug 2021
- 06:03 am
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.
- b.) Find the height of this uniform distribution.
- c.) Find the probability of waiting between four and five minutes.
- d.) Find the probability of waiting between three and eight minutes.
- e.) Find the probability of waiting five minutes exactly.
No data attached
6.3.2
Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal distribution with mean of m=0 and standard deviation s=1 .
a.) The area to the left of z is 15%.
Z-score = -1.036433
b.) The area to the right of z is 65%.
Z-Score = -0.3853205
c.) The area to the left of z is 10%.
Z-score= -1.281552
d.) The area to the right of z is 5%.
Z-Score = -1.644854
e.) The area between -z and z is 95%. (Hint draw a picture and figure out the area to the left
of the -z .)
Z-score: 1.959964, -Z = -1.959964
f.) The area between -z and z is 99%.
Z-Score = 2.575829, -Z = -2.575829
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.
The blood pressure measure is the random variable.
b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.
Probability = 0.3804315
c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.
Probability = 0.714037
d.) Find the probability that a person in China has blood pressure between 120 and 125 mmHg.
Probability= 0.08412583
e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?
No. The probability of having more than 135 mmHG is 0.38 which is very high to be unusual.
f.) What blood pressure do 90% of all people in China have less than?
90 th Percentile = 157.4757
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.
Life of dishwasher is the random variable here.
b.) Find the probability that a dishwasher will last more than 15 years.
Probability = 0.0008197
c.) Find the probability that a dishwasher will last less than 6 years.
Probability = 7.933282 * 10 -7
d.) Find the probability that a dishwasher will last between 8 and 10 years.
Probability = 0.054112
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. The probability of lasting 6 or less years is very small and hence considered unusual without any systematic reason.
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?
The lowest 5% Quantile is: 9.943933 ~ 9.5years
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.
The yearly rainfall in Sydney, Australia.
b.) Find the probability that the yearly rainfall is less than 100 mm.
Probability = 0.2958992
c.) Find the probability that the yearly rainfall is more than 240 mm.
Probability = 0.06775085
d.) Find the probability that the yearly rainfall is between 140 and 250 mm.
Probability = 0.431916
e.) If a year has a rainfall less than 100mm, does that mean it is an unusually dry year? Why or why not?
No. The probability of having less than 100mm rain is ~0.30 which is high for unusual event.
f.) What rainfall amount are 90% of all yearly rainfalls more than?
The 90 th Percentile is: = 225.4271
6.4.4
Annual rainfalls for Sydney, Australia are given in table #6.4.6. ("Annual maximums of," 2013). Can you
assume rainfall is normally distributed?
Table #6.4.6: Annual Rainfall in Sydney, Australia
146.8 383 90.9 178.1 267.5 95.5 156.5 180
90.9 139.7 200.2 171.7 187.2 184.9 70.1 58
84.1 55.6 133.1 271.8 135.9 71.9 99.4 110.6
47.5 97.8 122.7 58.4 154.4 173.7 118.8 88
84.6 171.5 254.3 185.9 137.2 138.9 96.2 85
45.2 74.7 264.9 113.8 133.4 68.1 156.4
Shapiro-Wilk normality test
data: xyz
W = 0.90297, p-value = 0.0008972
No. The data does not look normal and the test also doesn’t support the same
6.5.2
A random variable is normally distributed. It has a mean of 245 and a standard deviation of 21.
a.) If you take a sample of size 10, can you say what the shape of the distribution for the sample mean is? Why?
The mean will also be normally distributed as the random variable itself is normally distributed.
b.) For a sample of size 10, state the mean of the sample mean and the standard deviation of
the sample mean.
Mean = 245
Standard Deviation = SD10= 6.640783
c.) For a sample of size 10, find the probability that the sample mean is more than 241.
Probability = 0.7265256
d.) If you take a sample of size 35, can you say what the shape of the distribution of the sample mean is? Why?
The distribution of sample mean would still be normal as the random variable itself is normal.
e.) For a sample of size 35, state the mean of the sample mean and the standard deviation of the sample mean.
Mean = 245
Standard Deviation = SD35= 3.549648
f.) For a sample of size 35, find the probability that the sample mean is more than 241.
Probability = 0.8701
g.) Compare your answers in part c and f. Why is one smaller than the other?
The c part is smaller. Because as the sample size increases, mean of the sample clusters around mean of the random variable and the standard deviation decreases by factor of n
6.5.4
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). Blood pressure is normally distributed.
a.) State the random variable.
Blood Pressure of people in China
h.) Suppose a sample of size 15 is taken. State the shape of the distribution of the sample mean.
The mean will also be normally distributed as the random variable itself is normally distributed.
b.) Suppose a sample of size 15 is taken. State the mean of the sample mean.
Sample Mean = 128 mmHG
c.) Suppose a sample of size 15 is taken. State the standard deviation of the sample mean.
Standard Deviation =
d.) Suppose a sample of size 15 is taken. Find the probability that the sample mean blood pressure is more than 135 mmHg.
Probability = 0.0721999
e.) Would it be unusual to find a sample mean of 15 people in China of more than 135 mmHg? Why or why not?
The probability of having mean blood pressure of 15 people greater than 135 is greater than 0.05, hence not unusual.
f.) If you did find a sample mean for 15 people in China to be more than 135 mmHg, what might you conclude?
We may conclude that this is just by random chance and not due to any unusual sample.
6.5.6
The mean cholesterol levels of women age 45-59 in Ghana, Nigeria, and Seychelles is 5.1 mmol/l and the
standard deviation is 1.0 mmol/l (Lawes, Hoorn, Law & Rodgers, 2004). Assume that cholesterol levels
are normally distributed.
a.) State the random variable.
Cholesterol levels of women aged 45-59 in Ghana, Nigeria.
b.) Find the probability that a woman age 45-59 in Ghana has a cholesterol level above 6.2 mmol/l (considered a high level).
Probability = 0.1356661
c.) Suppose doctors decide to test the woman’s cholesterol level again and average the two values. Find the probability that this woman’s mean cholesterol level for the two tests is above 6.2 mmol/l.
Probability = 0.05989747
d.) Suppose doctors being very conservative decide to test the woman’s cholesterol level a third time and average the three values. Find the probability that this woman’s mean cholesterol level for the three tests is above 6.2 mmol/l.
Probability = 0.02837341
e.) If the sample mean cholesterol level for this woman after three tests is above 6.2 mmol/l, what could you conclude?
After three measurement, the probability of the event is ~0.028 which is less than 0.05 which makes it unusual event at 5% level of significance. Hence, it can be concluded that woman has unusually high level of cholesterol and the high measurement is not by
random error but may be a health problem.
6.5.8
A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance life expectancy," 2013). The life of a dishwasher is normally distributed. Suppose you are a manufacturer and you take a sample of 10 dishwashers that you made.
a.) State the random variable.
Life of a dishwasher in years.
b.) Find the mean of the sample mean.
Mean of sample mean = 12 years
c.) Find the standard deviation of the sample mean.
Standard Deviation of sample mean = 1.2510=0.3952847
d.) What is the shape of the sampling distribution of the sample mean? Why?
The distribution of sample mean would still be normal as the random variable itself is normal.
e.) Find the probability that the sample mean of the dishwashers is less than 6 years.
Probability = 2.438258 * 10 -52
f.) If you found the sample mean life of the 10 dishwashers to be less than 6 years, would you think that you have a problem with the manufacturing process? Why or why not?
The event is highly unusual, and I’d think that manufacturing process is flawed. It is very unusual to happen at random and there must be some systematic error.
