1.A population of values has a normal distribution with

=

231.8

=231.8

and

=

57.3

=57.3

. You intend to draw a random sample of sizen

=

73

n=73

.

FindP₆₅, which is the mean separating the bottom 65% means from the top 35% means.

P₆₅(for sample means) =

answers as numbers accurate to 1 decimal place. Answers obtained using exactz-scores orz-scores rounded to 3 decimal places are accepted.

2.A population of values has a normal distribution with

=

132.9

=132.9

and

=

75.5

=75.5

. You intend to draw a random sample of sizen

=

115

n=115

.

1. Find the probability that a single randomly selected value is greater than 142.1.P(X> 142.1) =Round to 4 decimal places.

Find the probability that the sample mean is greater than 142.1.P(X> 142.1) =Round to 4 decimal places.

3.A fitness company is building a 20-story high-rise. Architects building the high-rise know that women working for the company have weights that are normally distributed with a mean of 143 lb and a standard deviation of 29 lb, and men working for the company have weights that are normally distributed with a mean of 173 lb and a standard deviation or 28 lb. You need to design an elevator that will safely carry 15 people. Assuming a worst case scenario of 15 male passengers, find the maximum total allowable weight if we want a 0.98 probability that this maximum will not be exceeded when 15 males are randomly selected.

maximum weight =-lb Round to the nearest pound.

Answers obtained using exactz-scores orz-scores rounded to 2 decimal places

4.A population of values has a normal distribution with

=

133.2

=133.2

and

=

41

=41

. You intend to draw a random sample of sizen

=

174

n=174

.

Find the probability that a sample of sizen

=

174

n=174

is randomly selected with a mean between 138.8 and 140.3.

P(138.8 <M< 140.3) =

answers as numbers accurate to 4 decimal places. Answers obtained using exactz-scores orz-scores rounded to 3 decimal places

5.The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.883 g and a standard deviation of 0.289 g. The company that produces these cigarettes claims that it has now reduced the amount of nicotine. The supporting evidence consists of a sample of 42 cigarettes with a mean nicotine amount of 0.794 g.

Assuming that the given mean and standard deviationhave NOT changed, find the probability of randomly seleting 42 cigarettes with a mean of 0.794 gor less.

P(M< 0.794 g) =

answer as a number accurate to 4 decimal places. Answers obtained using exactz-scores orz-scores rounded to 3 decimal places are accepted.

Based on the result above, is it valid to claim that the amount of nicotine is lower?(Let's use a 5% cut-off for our definition of unusual.)

6.Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 169000 dollars. Assume the standard deviation is 39000 dollars. Suppose you take a simple random sample of 90 graduates.

Find the probability that a single randomly selected salary has a mean value between 172699.9 and 181744 dollars.

P(172699.9 <X< 181744) =(Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a random sample of sizen

=

90

n=90

has a mean value between 172699.9 and 181744 dollars.

P(172699.9 <

x

x

< 181744) =( answers as numbers accurate to 4 decimal places.)

7.A manufacturer knows that their items have a normally distributed length, with a mean of 5.8 inches, and standard deviation of 1.9 inches.

If 10 items are chosen at random, what is the probability that their mean length is less than 4.9 inches?

8.Express the confidence interval(

18.8

%

,

30.6

%

)

(18.8%,30.6%)

in the form of

p

M

E

p^ME

.

9.If n=110 and

p

p^

(p-hat) =0.21, find the margin of error at a 95% confidence level

answer to three decimals

10.115 students at a college were asked whether they had completed their required English 101 course, and 83 students said "yes". Construct the 90% confidence interval for the proportion of students at the college who have completed their required English 101 course.

Enter your answers as decimals (not percents) accurate to three decimal places.

The Confidence Interval is (,)