Download Assume the readings on thermometers are normally distributed with

1. Assume the readings on thermometers are normally distributed with a mean of 0
degrees C and a standard deviation of 1.00 degrees C. Find the probability that a
randomly selected thermometer reads between -2.18 and -1.52 and draw a sketch of
the region.
The probability is 0.0496 (rounded to 4 dp)
2) Assume that z scores are normally distributed with a mean of 0 and a standard
deviation of 1. (round all to 2 decimal places)
A. If P(zb)=0.9699, find b= Check this question please
C. if P(z>c)=0.0244, find c= 1.97
D. If P(-d<z<d)=0.5098, find d= 0.69
E. if P(-e<z<e)=0.0876, find e= 0.11
3) Women's heights are normally distributed with mean 63.9 in and standard
deviation of 2.5 in. A social or organization for tall people has a requirement that
women must be at least 70 in tall. What percentages of women meet that
requirement?
The percentage of women that are taller than 70 in is 0.73 % ( round to 2 decimal
places)
4) A survey found that women's heights are normally distributed with mean 63.4 in
and standard deviation 2.5 in. A branch of the military requires women's heights to
be between 58 in and 80 in.
A. Find the percentage of women meeting the heights requirement. Are many
women's being denied the opportunity to join this branch of the military because
they are too short or too tall?
B. If this branch of the military changes the height requirements so that all women
are eligible except the shortest 1% and the tallest 2% what are the new heights
requirements?
A) The percentage of women who meet the height requirements is 98.46 (round to 2
decimal places)
Are many women being denied the opportunity to join this branch of the military
because they are too short or too tall?
(Yes or No) why? Because women being denied are 1.54% (a small %)
B) For the new height requirements, this branch of the military requires women's
height to be at least 57.6 In and at most 68.5 In. (round to 1 decimal place)
5) The captivity of a lift is 8 people or 1376 pounds. The captivity will be exceeded
if 8 people have weights with a mean greater than 1376/8=172 pounds. Suppose the
people have weights that are normally distributed with a mean of 178 lb and
standard deviation of 34lb.
a. Find the probability that if a person is randomly selected, his weight will be
greater than 172 pounds. The probability is approximately 0.5700 (round to 4
decimal places)
b. Find the probability that 8 randomly selected people will have a mean that is
greater than 172 pounds. The probability is approximately 0.6912. (round to 4
decimal places)
c. Does the lift appear to have the correct weight limit? Why or why not?
No, because the probability that 8 people exceeds the limit is not small (0.6912)
6. Case of a cretin beverage are labeled to indicate that they contain 16 oz. The
amounts in a sample of cans are measured and the sample statistic are n= 39 and
x=16.01 oz. If the beverage cans are filled so that u=16.00 oz ( as labeled) and the
population standard deviation is o=0.149 oz. (based on the sample results), find the
probability that a sample of 39 cans will have a mean of 16.01 oz or greater. Do
these results suggest that the beverage cans are filled with an amount greater than
16.00 oz?
a. The probability that a sample of 39 cans will have a mean of 16.01 oz or greater,
given that u=16.00 o-0.149, is 0.3376 .( round to 4 decimal places)
b. Do these results suggest that the beverage cans are filled with an amount greater
than 16.00 oz? No, because the probability that a sample of 39 cans will have a
mean of 16.01 oz or greater given that u=16.00 is considerable
7. Use the given margin of error, confidence level, and population standard
deviation, o, to find the minimum sample size required to estimate an unknow
population mean, u.
Margin of error: 1.8 inches, confidence level: 95% o=2.6 inches
The confidence level of 95% requires a minimum sample size of 8 . ( round to the
nearest integer)
8. The data shown below results from using a random sample of speeds of drivers
ticketed on a section of an interstate. Using the display, identify the value of the
point estimate of the population mean u.
Variable---N—Mean—StDev-----SE Mean-----------95% CI
Speed ---73--65.514--3.7532----0.4393 ----------(64.653,66.375)
The point estimate of the population mean u is 65.514.
9. Using the simple random sample of weights of women from a data set, we obtain
these sample statistics: n=45 and x=147.84 lb. Research from other sources suggests
that the population of weights of women has a standard deviation given by o=30.55
lb.

a. Find the best point estimate of the mean weight of all women.
The best point estimate is 147.84 lb.

b. Find a 90% confidence interval estimate of the mean weight of all women.
The 90% confidence interval estimate is 140.35 lb < u < 155.33 lb.
10. Salaries of 44 college graduates who took a statistics course in college have a mean, x,
of $61,700. Assuming a standard deviation, o, of $11,759, construct a 90% confidence
interval for estimating the population mean u.
$ 58784 < u < $ 64616 (round to the nearest integer as needed)
11. In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were tested
with garlic in a processed tablet form. Cholesterol levels were measured before and after
the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 3.8
and a standard deviation of 19.8. Complete parts (a) and (b) below.

a. What is the best point estimate of the population mean net change in LDL
cholesterol after the garlic treatment?
The best point estimate is 3.8 mg/dL.

b. Construct a 99% confidence interval estimate of the mean net change in LDL
cholesterol after the garlic treatment. What does the confidence interval suggest
about the effectiveness of garlic in reducing LDL cholesterol?
What is the confidence interval estimate of the population mean u?
-3.4 mg/dL < u < 11.0 mg/dL ( round to 1 decimal place as needed)

c. What does the confidence interval suggest about the effectiveness of the
treatment?
The effectiveness is not good, since 0 is in the interval we can´t decide that the
treatment has a positive mean (change is positive)
12. A random sample of the birth weights of 186 babies has a mean of 3103 g and a
standard deviation of 692 g. construct a 95% confidence interval estimate of the mean birth
weight for all babies.
What is the confidence interval estimate of the mean birth weight for such babies?

a. 3004 g<u< 3202 ( round to nearest integer as needed)
13. In a survey of 155 senior executives 51% said that most common job interview mistake
is to have little or no knowledge of the company. Test the claim that is the population of all
senior executives, 40% say that the most common job interview mistake is to have little or
no knowledge of the company. Identify the null hypothesis, alternative hypothesis, test
statistic, p-value, conclusion and the null hypothesis, and final conclusion that address the
original claim. Use the p-value method. Use the normal disruption as an approximation of
the binomial distribution.

b. Identify the null value and alternative hypotheses.
Null: Ho: p = 0.4
Alternative: Ha: p ≠ 0.4



c. The test statistic is z = 2.80 ( round to 2 decimal places as needed)
d. The p-value is 0.0052 ( round to 4 decimal places as needed)
e. Identify the conclusion about the null hypothesis and the final conclusion that
addresses the original claim. (assume a 0.05 significance level)
Since the p-value = 0.0052 < 0.05 we reject Ho
14. Test the following claim. Identify the null hypothesis, alternative hypothesis test
statistic, critical values(s), conclusion about the null hypothesis, and final conclusion that
addresses the original claim.
A simple random sample of 40 salaries of professional football coaches has a mean of
$415,945. The standard deviation of all salaries of professional coaches is $466,613. Use a
0.01 significance level to test the claim that the mean salary of professional football coach
is less than $500,000.

a. What are the null and alternative hypotheses?
Null: Ho:  ≥ 500,000
Alternative: Ha:  < 500,0000

b. What is the value of the test statistic?
Z = -1.14 (round to two decimal places)

c. Identify the critical value(s) of z.
Z= -2.33 (round to two decimal places)
d. Is there sufficient evidence to support the claim that the mean salary of a professional
football coach is less than $500,000? No, because the statistic value (-1.14) is greater than
the critical value (-2.33)
15. Assume that a simple random sample has been selected from a normally distributed
population and test the given claim. Identify the null and alternative hypotheses, test
statistic, p-value, critical value(S), and state the final conclusion that addresses the original
claim.
A safety administrator conducted crash test of child booster seats for cars, Listed below are
results from those tests, with the measurements given in hic (standard head injury condition
units). The safety requirements is that the hic measurement should be less than 1000 hic.
Use a 0.01 significance level to test the claim that the sample is from a population with a
mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the
specified requirement?
749 , 565, 1102, 598, 605, 568



a. What are the hypotheses?
Null: Ho:  ≥ 1,000
Alternative: Ha:  < 1,0000




b. Identify the test statistic. t = -3.536 (round to 3 decimal places)
c. Identify the p-value. 0.0083 (round to 4 decimal places)
d. Identify the critical value(s) -3.365 (round to 3 decimal places)
e. State the final conclusion that addresses the original claim.
Since the p-value (0.0083) is less than 0.01 we reject Ho
What do the results suggest about the child booster seats meeting the specified
requirement?




a. There is not strong evidence that the mean is less than 1000 hic, and one of the
booster seats has a measurement that is greater than 1000 hic
b. There is strong evidence that the mean is less than 1000 hic, but one of the
booster seats has a measurement that is greater than 1000 hic
c. The requirement is met since most sample measurements are less than 1000 hic.
d. The results are inconclusive regarding whether one of the booster seats could
have a measurement that is greater than 1000 hic.
16. The data show the chest size and weight of several bears. Find the regression equation,
letting the first variable be the independent(x) variable. Then find the best predicted weight
of a bear with a chest size of 39 inches. Use a significance level of 0.05.
Chest size (inches) 38, 41,49,45,35,48,
Weight (pounds) 319, 345,531,451,326,463

a. What is the regression equation?
Y= -223.77 + 14.76 x (round to two decimal places)

b. What is the best predicted value?
Y= 351.87 (round to two decimal places)