Review Ch.7-9.1

Verify that the requirements for constructing a confidence interval about 𝑝 are satisfied.

np(1-p)≥10 n<0.05N Data comes from simple random sample

The random number generator on a calculator randomly generates a number between 0 and 1. The random variable 𝑋, the number generated, follows a uniform probability distribution. What is the probability of generating a number between 0 and 0.2?

(.2-0)•1=.2

What is the probability of generating a number between 0.25 and 0.6?

(.6-.25)•1=.35

What is the probability of generating a number greater than 0.95?.

(1-.95)•1=.05

What is the probability that the reaction time is between 5 and 8 minutes?

(8-5)•1/5=.6

2. The reaction time 𝑋 (in minutes) of a certain chemical process follows a uniform probability distribution with 5 ≤ 𝑋 ≤ 10. What is the probability that the reaction time is between 6 and 8 minutes?

(8-6)•1/5=.4

Mean length 25cm standard deviation=0.07 B)=.0322 Using the results of part b, if 5000 rods are manufactured in a day, how many should the plant manager expect to discard?

.0322•5000=161

Verify that the requirements for constructing a confidence interval about 𝑝 are satisfied. See work from class

np(1-p)≥10 n<0.05N Data comes from simple random sample

A random sample of 1003 adult Americans was asked, "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1003 adults surveyed, 521 indicated that televisions are a luxury they could do without. a. Obtain a point estimate for the population proportion of adult Americans who believe that televisions are a luxury they could do without.

521/1003=.52

Would it be unusual for a random sample of 500 adults to result in 125 or fewer who do not own a credit card? Why?

125/500=.25 normalcdf(-99999,.25,,29,.02)=.02 yes would be ununusal

Suppose a popular flight has 300 seats and 300 reservations. Because this flight is popular, the proportion of passengers with a reservation who miss the flight is only 0.04. You are booked on a later flight and put yourself on the stand-by list. There are 14 passenger names ahead of you. What is the probability you get on this flight?

15/300=.05 normalcdf(.05,99999,.04,√(.4•.96)/300=0.18

Suppose a flight has 320 reservations but only 300 seats on the plane. What is the probability that 300 or fewer passengers show up for the flight?

300/320=15/16 normalcdf(-99999,15/16,.9005,√(.0995•.9005)/320=.99

A survey of 2306 adult Americans conducted by Harris Interactive found that 417 have donated blood in the past two years. a. Obtain a point estimate for the population proportion of adult Americans who have donated blood in the past two years.

417/2306=.181

4. Assume that the random variable 𝑋 is normally distributed, with mean 𝜇 = 50 and standard deviation 𝜎 = 7. Find the indicated percentile for 𝑋. The 9th percentile

InvNorm(.09,50,7)=40.62

Treat the next 50 cars as a simple random sample. What is the 15th percentile of the sample mean?

InvNorm(.15,59,13,1/√50)=57.4

mean 𝜇 = 50 and standard deviation 𝜎 = 7. Find the indicated percentile for 𝑋. The 38th percentile

InvNorm(.38,50,7)=47.86

mean 𝜇 = 50 and standard deviation 𝜎 = 7. Find the indicated percentile for 𝑋. The 81st percentile

InvNorm(.81,50,7)=56.15

mean 𝜇 = 50 and standard deviation 𝜎 = 7. Find the indicated percentile for 𝑋. The 90th percentile

InvNorm(.90,50,7)=58.97

Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil change facility will perform 40 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the manager.

InvNorm=(.1,0,1)=-1.28 11.4+(-1.28)(3.2/√40)=10.8 minutes

Is it possible that a supermajority (more than 60%) of adult Americans believe that television is a luxury they could do without? Is it likely?

Yes it possible though it is not likely since p(P hat≥.6)≤p( phat≥0.55<.025<0.05)

What might you conclude if a random sample of 50 pregnancies resulted in a mean gestation period of 260 days or less? .004

Yes it would be considered unusally beacuase .004<.05

Construct and interpret a 95% confidence interval for the population proportion of adult Americans who believe that televisions are a luxury they could do without.

Z0.025=1.96√(.52•.48)/1003=.03 .52-0.3=.49 .52+3=.55 (.49,.55)

c. Construct a 90% confidence interval for the population proportion of adult Americans who have donated blood in the past two years. ( (𝟎. 𝟏𝟕, 𝟎. 𝟏𝟗) ) d. Interpret this interval.

Z0.05√.181(1-.181)/2306=1.645•.00808=.01 P-e p+e .01+.18=.19 .18-.01=.17 (.17,.19) We are 90% confidence that the population proportion of adults that donte blood is between 17% and 19%

What sample size should be obtained if she wishes the estimate to be within 4 percentage points with 90% confidence if She does not use any prior estimates? (423) c. Why are the results from part a and part b so close? (𝟎. 𝟓𝟓 ⋅ 𝟎. 𝟒𝟓 ≅ 𝟎. 𝟐𝟓)

n=0.25•(1.645/.04)=423 (0.55)(.45)=.25

10 min oil change However, records indicate that the mean time for an oil change is 11.4 minutes and the standard deviation for oil change time is 3.2 minutes. What is the probability that a random sample of 𝑛 = 40 oil changes results in a sample mean time of less than 10 minutes?

normalcdf(-99999,11.4,3.2/√40)=.003

59.3 seconds with a standard deviation of 13.1. The quality-control manager wishes to use a new delivery system designed to get cars through the drive-through system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. What is the probability of obtaining a sample mean of 56.8 seconds or less, assuming that the population mean is 59.3 seconds? Do you think this new system is effective?

normalcdf(-99999,56.8,59.3,13.1/√40)=0.1137 This system is not effective P(x bar≤56.8)>0.05

Steel rods are manufactured with a mean length of 25 centimeters (cm). Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed, with a standard deviation of 0.07 cm. What proportion of rods has a length less than 24.9 cm?

normalcdf(-999999,24,9,25,.07)=.08

6. The length of human pregnancies is approximately normally distributed with mean 𝜇 = 266 days and standard deviation 𝜎 = 16 days. What is the probability a randomly selected pregnancy lasts less than 260 days?

normalcdf(-999999,260,266,16)=.35

𝜇 = 266 days What is the probability that a random sample of 50 pregnancies resulted in a mean gestation period of 260 days or less?

normalcdf(-999999,260,266,𝟏𝟔/√50)=.004

𝜇 = 266 days and standard deviation 𝜎=𝟏𝟔/√𝟐𝟎 days What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

normalcdf(-999999,260,266,𝟏𝟔/√𝟐𝟎)=.05

What is the probability that in a random sample of 500 adults between 25% and 30% do not own a credit card?

normalcdf(.25,.3,29,.02)=.66

What is the probability that in a random sample of 500 adults more than 30% do not own a credit card?

normalcdf(.3,99999,29,.02)=.31111

Mean length 25cm standard deviation=0.07 Any rods that are shorter than 24.85 cm or longer than 25.15 cm are discarded. What proportion of rods will be discarded?

normalcdf(24.85,25.15,25,.07)=.9678 1-.9678=.0322

Mean length 25cm standard deviation=0.07 If an order comes in for 10,000 steel rods, how many rods should the plant manager manufacture if the order states that all rods must be between 24.9 cm and 25.1 cm.

normalcdf(24.9,25,1,25,0.07)=.8468 .8468x=10000 x=11800 About

What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

normalcdf(256,276,266,𝟏𝟔/√15)=.98

3. Assume that that the random variable 𝑋 is normally distributed, with mean 𝜇 = 50 and standard deviation 𝜎 = 7. Compute the following probabilities. Be sure to draw a normal curve with the area corresponding to the probability shaded. 𝑃(𝑋 > 35)

normalcdf(35,99999,50,7)=.98

mean 𝜇 = 50 and standard deviation 𝜎 = 7. Compute the following probabilities. 𝑃(38 < 𝑋 ≤ 55)

normalcdf(38,55,50,7)=.72

mean 𝜇 = 50 and standard deviation 𝜎 = 7. Compute the following probabilities. 𝑃(56 ≤ 𝑋 <66)

normalcdf(56,66,50,7)=.18

mean 𝜇 = 50 and standard deviation 𝜎 = 7. Compute the following probabilities. 𝑃(𝑋 > 65)

normalcdf(65,99999,50,7)=.02

mean 𝜇 = 50 and standard deviation 𝜎 = 7. Compute the following probabilities. 𝑃(𝑋 ≤ 45)

normalcdf(99999,45,50,7)=.24

The quality-control manager of a 7 leaves coffee shop wants to analyze the length of wait time that a car spends at the drive-through window waiting for an order. It is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1. The distribution of time spent at the window is skewed right. To obtain probabilities regarding a sample mean using the normal model, what size sample is required?

n≥30

The shade of the distribution of the time required to get an oil change at a 10-minute oil change facility is unknown. However, records indicate that the mean time for an oil change is 11.4 minutes and the standard deviation for oil change time is 3.2 minutes. To compute probabilities regarding the sample mean using the normal model, what size sample would be required?

n≥30

A sociologist wishes to conduct a poll to estimate the percentage of Americans who favor affirmative action programs for women and minorities for admission to colleges and universities. What sample size should be obtained if she wishes the estimate to be within 4 percentage points with 90% confidence if a. She uses a 2003 estimate of 55% obtained from a Gallup Youth Survey?

p hat(1-p hat)•(Zα/2//E=.55•.45(1.645/.04)=419

Let's assume that the proportion of passengers who miss a flight for which they have a reservation is 0.0995. a. Suppose a flight has 290 reservations. What is the probability that 25 or more passengers will miss the flight?

p hat=25/290=.86 normalcdf(.86,99999,√(.0995•.9005)/290=.78

𝜇 = 266 days and standard deviation 𝜎 = 16 days Suppose a random sample of 20 pregnancies is obtained. Describe the sampling distribution of the sample mean length of human pregnancies.

𝜇x = 266 days 𝜎x=16/√20

Suppose 29% of adults do not own a credit card. a. Suppose a random sample of 500 adults are asked "Do you own a credit card?" Describe the sampling distribution of 𝑝̂, the proportion of adults who do not own a credit card.

𝝁𝒑̂ = 𝟎. 𝟐𝟗, 𝝈𝒑̂ = 𝟎. 𝟎𝟐) σp=√.29(1-.29)/500=.0203