SPEA-V 506 Exam Two (Chapters 4-8)

For some value of Z, the probability of the cumulative standardized normal distribution is 0.2090. The value of Z is

-0.81, use =NORM.INV(probability, mean, std dev). So =NORM.INV(0.2090, 0, 1)

The collection of all possible events is called

sample space

The t distribution approaches the standardized normal distribution when the number of degrees of freedom increases.

True

A probability distribution

associates a probability of occurrence with each outcome.

A sampling distribution is a distribution for a statistic.

True

A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Prices for 100 rats follow the following distribution: How much should the lab budget for next year's rat orders be, assuming this distribution does not change?

$637, multiply the expected value by the number of weeks. Use =SUMPRODUCT to find $12.625, multiply by 52

As the sample size increases, the standard error of the mean increases.

False

In estimating the population mean with the population standard deviation unknown, if the sample size is 12, there will be 6 degrees of freedom.

False

The probability that a standard normal variable, Z, is below 1.96 is 0.4750.

False

In its standardized form, the normal distribution

has a mean of 0 and a standard deviation of 1

A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit to the chain's new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: Sample mean = $50.50 and Sample std dev = 20 . Construct a 95% confidence interval for the mean amount its credit card customers spent on their first visit to the chain's new store in the mall if the amount spent follows a normal distribution.

$50.50 ± $11.08, 1. Find Critical Value =ABS(NORM.S.INV(alpha/2)) 2. Calculate Standard Error 3. Calculate Margin of Error - Margin of Error = Critical value * Standard Error 4. Create lower limit by subtracting Margin of error from mean 5. Create upper limit by adding Margin of error to mean

An alcohol awareness task force at a Big-Ten university sampled 200 students after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following result was obtained. What is the probability that a randomly selected student did well on the midterm or went bar hopping the weekend before the midterm?

(80+30+70)/200 or (110+100-30)/200 or 90%

If two events are mutually exclusive, what is the probability that both occur at the same time?

0

A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients would receive payments beyond age 75? Round your answer to 4 decimals.

0.0228 (with margin: 0) In excel use the formula =1-NORM.DIST

At a computer manufacturing company, the actual size of a computer chip is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the standard error for the sample mean?

0.029, to find standard error, divide the standard deviation by the square root of n. =0.1/SQRT(12)

The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh more than 4.4 pounds is... round your answer to 4 decimals

0.0668 (with margin: 0) In excel, use the formula =1-NORM.DIST (4.4, 3.2, 0.8, TRUE)

The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 16 fish is taken, what would the standard error of the mean weight equal?

0.200, to find standard error, divide the standard deviation by the square root of n =0.8/SQRT(16)

If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes.

0.3085, use =NORM.DIST(3, 3.5, 1, TRUE)

The probability of the cumulative standardized normal distribution at Z is 0.6255. The value of Z is

0.32

The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 1024K data package through a network. X0123P(X)0.350.350.250.05 The probability of no retransmissions is

0.35

The amount of tea leaves in a can from a production line is normally distributed with mean of 110 grams and standard deviation of 25 grams. What is the probability that a randomly selected can will contain at least 100 grams of tea leaves? round your answer to 4 decimals.

0.6554 (with margin: 0) In excel: use formula norm.dist (x, mean, std dev, true). For this problem, we will use 1-norm.dist because it says "at least". Enter =1-NORM.DIST( 100, 110, 25, TRUE)

The employees of a company were surveyed on questions regarding their educational background (college degree or no college degree) and marital status (single or married). Of the 600 employees, 400 had college degrees, 100 were single, and 60 were single college graduates. The probability that an employee of the company is single or has a college degree is:

0.733, use P(A or B=P(A)+P(B)-P(A intersect B), so 400+100-60=440, 440/600=0.733

If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot.

0.7745, use =NORM.DIST(4.5, 3.5, 1)-NORM.DIST(2, 3.5, 1)

The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 1024K data package through a network. X0123P(X)0.350.350.250.05 Calculate the variance for the number of retransmissions. Round to 2 decimals.

0.8, use the dispersion formula to find variance. Insert data into excel 1.Calculate Expected Value (Mean) 2.Create Squared Deviation column =(Value - Mean)2 4.Multiple Squared Deviation column by probability 5.Sum column created in step 4 to find the Variance 6.Take the square root of the Variance to find the Standard Deviation

The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be less than 124 inches? Round your answer to 4 decimals.

0.8413 (with margin: 0)

The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 1024K data package through a network. X0123P(X)0.350.350.250.05 Calculate the standard deviation for the number of retransmissions. Round to 3 decimals.

0.894 (with margin: 0.001) In Excel: 1.Calculate Expected Value (Mean) 2.Create Squared Deviation column =(Value - Mean)2 4.Multiple Squared Deviation column by probability 5.Sum column created in step 4 to find the Variance 6.Take the square root of the Variance to find the Standard Deviation

Given that X is a normally distributed variable with a mean of 50 and a standard deviation of 2, find the probability that X is between 47 and 54. Round your answer to 4 decimals.

0.9104 In excel, use NORM.DIST to find both x values and the subtract the larger number and the smaller number. =NORM.DIST(54, 50, 2, TRUE)-NORM.DIST(47, 50, 2, TRUE)

The following table contains the probability distribution for X = the number of retransmissions necessary to successfully transmit a 1024K data package through a network. X0123P(X)0.350.350.250.05 Calculate the mean or expected value for the number of retransmissions

1

The probability of the cumulative standardized normal distribution at Z is 0.8770. The value of Z is

1.16, use =NORM.INV

The amount of time required for an oil and filter change on an automobile is normally distributed with a mean of 45 minutes and a standard deviation of 10 minutes. A random sample of 16 cars is selected. What is the standard error of the mean? round your answer to 1 decimals.

2.5

A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine. The desired mean length of the insulation is 12 feet. It is known that the standard deviation in the cutting length is 0.15 feet. A sample of 70 cut sheets yields a mean length of 12.14 feet. This sample will be used to obtain a 99% confidence interval for the mean length cut by machine. The critical value to use in obtaining the confidence interval is

2.58

If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, 75.8% of the college students will take more than how many minutes when trying to find a parking spot in the library parking lot?

2.8 minutes, =NORM.INV (probability below, mean, std dev) so =NORM.INV(0.242, 3.5, 1)

A business venture can result in the following outcomes (with their corresponding chance of occurring in parentheses): Highly Successful (10%), Successful (25%), Break Even (25%), Disappointing (20%), and Highly Disappointing (?). If these are the only outcomes possible for the business venture, what is the chance that the business venture will be considered Highly Disappointing?

20%

An alcohol awareness task force at a Big-Ten university sampled 200 students after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following result was obtained. What is the probability that a randomly selected student who went bar hopping did well on the midterm?

30/100 or 30%

An alcohol awareness task force at a Big-Ten university sampled 200 students after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following result was obtained. 80 20 30 70 What is the probability that a randomly selected student did well on the midterm and went bar hopping the weekend before the midterm?

30/200 or 15%

Mothers Against Drunk Driving is a very visible group whose focus is to educate the public about the harm caused by drunk drivers. A study was recently done that emphasized the problem we all face with drinking and driving. Four hundred accidents that occurred on a Saturday night were analyzed. Two items noted were the number of vehicles involved and whether alcohol played a role in the accident. The numbers are shown below: What proportion of accidents involved more than one vehicle?

325/400 or 81.25%

The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight (in pounds) should the citation designation be established? round your answer to 2 decimals.

4.48, =NORM.INV(0.98, 3,2, 0.8)

Mothers Against Drunk Driving is a very visible group whose focus is to educate the public about the harm caused by drunk drivers. A study was recently done that emphasized the problem we all face with drinking and driving. Four hundred accidents that occurred on a Saturday night were analyzed. Two items noted were the number of vehicles involved and whether alcohol played a role in the accident. The numbers are shown below: Given alcohol was involved, what proportion of accidents involved a single vehicle?

50/170 or 29.41%v

A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. Find the age at which payments have ceased for approximately 86% of the plan participants. Round your answer to 2 decimal places.

71.78 In excel, use the formula =NORM.INV(probability, mean, std dev)

If the outcome of event A is not affected by event B, then events A and B are said to be..

Independent

A lab orders 100 rats a week for each of the 52 weeks in the year for experiments that the lab conducts. Suppose the mean cost of rats used in lab experiments turned out to be $13.00 per week. Interpret this value.

The expected or mean cost for all weekly rat purchases is $13.00.

Another name for the mean of a probability distribution is its expected value.

True

The confidence interval estimate of the population mean is constructed around the sample mean.

True

The sample mean is a point estimate of the population mean.

True

The standard error of the mean is also known as the standard deviation of the sampling distribution of the sample mean.

True

True or False: A sample size of 5 provides a sample mean of 9.6. If the population variance is known to be 5 and the population distribution is assumed to be normal, the lower limit for a 90% confidence interval is 7.96.

True

The probability that a standard normal variable, Z, falls between - 1.50 and 0.81 is 0.7242.

True, use =NORMSDIST(z) for both and subtract the two

It is desired to estimate the mean total compensation of CEOs in the Service industry. Data were randomly collected from 18 CEOs and the 95% confidence interval was calculated to be ($2,181,260, $5,836,180). Which of the following interpretations is correct?

We are 95% confident that the mean total compensation of all CEOs in the Service industry falls in the interval $2,181,260 to $5,836,180.

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. The 95% confidence interval for π is 0.59 ± 0.07. Interpret this interval.

We are 95% confident that the true proportion of all students receiving financial aid is between 0.52 and 0.66.