Exam 6: The Normal Distribution

SCENARIO 6-2 John has two jobs.For daytime work at a jewelry store he is paid $15,000 per month, plus a commission.His monthly commission is normally distributed with mean $10,000 and standard deviation $2000.At night he works occasionally as a waiter, for which his monthly income is normally distributed with mean $1,000 and standard deviation $300.John's income levels from these two sources are independent of each other. -Referring to Scenario 6-2, for a given month, what is the probability that John's commission from the jewelry store is between $5,000 and $7,000?

You were told that the mean score on a statistics exam is 75 with the scores normally distributed.In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%.What is the probability of a score between 60 and95?

SCENARIO 6-2 John has two jobs.For daytime work at a jewelry store he is paid $15,000 per month, plus a commission.His monthly commission is normally distributed with mean $10,000 and standard deviation $2000.At night he works occasionally as a waiter, for which his monthly income is normally distributed with mean $1,000 and standard deviation $300.John's income levels from these two sources are independent of each other. -Referring to Scenario 6-2, the probability is 0.45 that John's income as a waiter is more than how much in a given month?

SCENARIO 6-4 According to Investment Digest, the arithmetic mean of the annual return for common stocks over an 85-year period was 9.5% but the value of the variance was not mentioned.Also 25% of the annual returns were below 8% while 65% of the annual returns were between 8% and 11.5%.The article claimed that the distribution of annual return for common stocks was bell-shaped and approximately symmetric.Assume that this distribution is normal with the mean given above.Answer the following questions without the help of a calculator, statistical software or statistical table. -Referring to Scenario 6-4, find the probability that the annual return of a random year will be more than 7.5%.

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

SCENARIO 6-5 Ball bearings are manufactured with a mean diameter of 6 millimeters (mm).Because of the inherent manufacturing process variability, the lots of bearings are approximately normally distributed with a standard deviation of 0.03 mm. -Using Scenario 6-5, what proportion of ball bearings has a diameter of greater than 6 mm? NEW QUESTION

The amount of time necessary for assembly line workers to complete a product is a normal variable with a mean of 15 minutes and a standard deviation of 2 minutes.So, 60% of the products would be assembled within and minutes (symmetrically distributed about the mean).

SCENARIO 6-4 According to Investment Digest, the arithmetic mean of the annual return for common stocks over an 85-year period was 9.5% but the value of the variance was not mentioned.Also 25% of the annual returns were below 8% while 65% of the annual returns were between 8% and 11.5%.The article claimed that the distribution of annual return for common stocks was bell-shaped and approximately symmetric.Assume that this distribution is normal with the mean given above.Answer the following questions without the help of a calculator, statistical software or statistical table. -Referring to Scenario 6-4, find the probability that the annual return of a random year will be between 7.5% and 11%.

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?

Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1.The probability that Z is between -0.88 and 2.29 is .

SCENARIO 6-2 John has two jobs.For daytime work at a jewelry store he is paid $15,000 per month, plus a commission.His monthly commission is normally distributed with mean $10,000 and standard deviation $2000.At night he works occasionally as a waiter, for which his monthly income is normally distributed with mean $1,000 and standard deviation $300.John's income levels from these two sources are independent of each other. -Referring to Scenario 6-2, John's commission from the jewelry store will be between what two values symmetrically distributed around the population mean 80% of the time?

You were told that the mean score on a statistics exam is 75 with the scores normally distributed.In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%.The middle 86.64% of the students will score between which two scores?

SCENARIO 6-2 John has two jobs.For daytime work at a jewelry store he is paid $15,000 per month, plus a commission.His monthly commission is normally distributed with mean $10,000 and standard deviation $2000.At night he works occasionally as a waiter, for which his monthly income is normally distributed with mean $1,000 and standard deviation $300.John's income levels from these two sources are independent of each other. -Referring to Scenario 6-2, for a given month, what is the probability that John's income as a waiter is between $700 and $1600?

SCENARIO 6-2 John has two jobs.For daytime work at a jewelry store he is paid $15,000 per month, plus a commission.His monthly commission is normally distributed with mean $10,000 and standard deviation $2000.At night he works occasionally as a waiter, for which his monthly income is normally distributed with mean $1,000 and standard deviation $300.John's income levels from these two sources are independent of each other. -Referring to Scenario 6-2, John's income as a waiter will be between what two values symmetrically distributed around the population mean 80% of the time?

SCENARIO 6-2 John has two jobs.For daytime work at a jewelry store he is paid $15,000 per month, plus a commission.His monthly commission is normally distributed with mean $10,000 and standard deviation $2000.At night he works occasionally as a waiter, for which his monthly income is normally distributed with mean $1,000 and standard deviation $300.John's income levels from these two sources are independent of each other. -Referring to Scenario 6-2, for a given month, what is the probability that John's income as a waiter is more than $900?

SCENARIO 6-3 A company producing orange juice buys all its oranges from a large orange orchard.The amount of juice that can be squeezed from each of these oranges is approximately normally distributed with a mean of 4.7 ounces and some unknown standard deviation.The company's production manager knows that the probability is 30.85% that a randomly selected orange will contain less than 4.5 ounces of juice.Also, the probability is 10.56% that a randomly selected orange will contain more than 5.2 ounces of juice.Answer the following questions without the help of a calculator, statistical software or statistical table. -Referring to Scenario 6-3, what is the probability that a randomly selected orange will contain at least 4.9 ounces of juices?

You were told that the mean score on a statistics exam is 75 with the scores normally distributed.In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%.What is the probability of a score between 55 and90?

SCENARIO 6-1 The number of column inches of classified advertisements appearing on Mondays in a certain daily newspaper is normally distributed with population mean of 320 and population standard deviation of 20 inches. -Referring to Scenario 6-1, for a randomly chosen Monday, what is the probability there will be between 280 and 360 column inches of classified advertisement?

The amount of time necessary for assembly line workers to complete a product is a normal variable with a mean of 15 minutes and a standard deviation of 2 minutes.So, 15% of the products require more than minutes for assembly.

The amount of tea leaves in a can from a production line is normally distributed with $\mu = 110$ grams and $\sigma = 25$ grams.What is the probability that a randomly selected can will contain lessthan 100 grams or more than 120 grams of tea leaves?