Exam 6: The Normal Distribution

A worker earns $15 per hour at a plant in China and is told that only 2.5% of all workers make a higher wage.If the wage is assumed to be normally distributed and the standard deviation of wage rates is $5 per hour, the mean wage for the plant is $7.50 per hour.

You were told that the amount of time elapsed between consecutive trades on a foreign stock exchange market followed a normal distribution with a mean of 15 seconds.You were also told that the probability that the time elapsed between two consecutive trades to fall between 16 to 17 seconds was 13%.The probability that the time elapsed between two consecutive trades would fall below 13 seconds was 7%.What is the probability that the time elapsed between two consecutive trades will be between 15 and 16 seconds?

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 standarddeviation of 3.5 years.What proportion of the plan recipients would receive payments beyond age75?

You were told that the amount of time elapsed between consecutive trades on a foreign stock exchange market followed a normal distribution with a mean of 15 seconds.You were also told that the probability that the time elapsed between two consecutive trades to fall between 16 to 17 seconds was 13%.The probability that the time elapsed between two consecutive trades would fall below 13 seconds was 7%.What is the probability that the time elapsed between two consecutive trades will be longer than 17 seconds?

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 6-5, what is the probability that a randomly selected orange will contain between 4.5 and 5.2 ounces of juices?

In its standardized form, the normal distribution

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 less than 11.5%.

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 90% of the time?

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 no more than 4.2 ounces of juices?

Any set of normally distributed data can be transformed to its standardized form.

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 more than $9,500?

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 $800 and $900?

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.The probability isthat a product is assembled in between 10 and 12 minutes.

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).

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 6-4 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, what is the value above which will account for the highest 25% of the possible annual returns?

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 between 121 and 124 inches?

You were told that the amount of time elapsed between consecutive trades on a foreign stock exchange market followed a normal distribution with a mean of 15 seconds.You were also told that the probability that the time elapsed between two consecutive trades to fall between 16 to 17 seconds was 13%.The probability that the time elapsed between two consecutive trades would fall below 13 seconds was 7%.The middle 60% of the time elapsed will fall between which two numbers?

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 two values that will bound the middle 80% of the annual returns?

The probability that a standard normal variable, Z, is between 1.00 and 3.00 is0.1574.