Exam 6: The Normal Distribution and Other Continuous Distributions

Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is more than -0.98 is ________.

TABLE 6-3 Suppose the time interval between two consecutive defective light bulbs from a production line has a uniform distribution over an interval from 0 to 90 minutes. -Referring to Table 6-3, what is the probability that the time interval between two consecutive defective light bulbs will be at least 50 minutes?

TABLE 6-6 According to Investment Digest, the arithmetic mean of the annual return for common stocks from 1926-2010 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 Table 6-6, find the probability that the annual return of a random year will be less than 7.5%.

The amount of time between successive TV watching by first graders follows an exponential distribution with a mean of 10 hours. The probability that a given first grader spends more than 5 hours between successive TV watching is ________.

TABLE 6-3 Suppose the time interval between two consecutive defective light bulbs from a production line has a uniform distribution over an interval from 0 to 90 minutes. -Referring to Table 6-3, what is the probability that the time interval between two consecutive defective light bulbs will be less than 10 minutes?

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 90 and 95?

TABLE 6-6 According to Investment Digest, the arithmetic mean of the annual return for common stocks from 1926-2010 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 Table 6-6, 75% of the annual returns will be lower than what value?

Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than 1.15 is ________.

The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is ________ that a product is assembled in more than 11 minutes.

Suppose the probability of producing a defective light bulb from a production line is the same over an interval of 90 minutes. Which of the following distributions would you use to determine the probability that a defective light bulb will be produced in a 15 minutes interval?

TABLE 6-6 According to Investment Digest, the arithmetic mean of the annual return for common stocks from 1926-2010 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 Table 6-6, find the probability that the annual return of a random year will be between 7.5% and 11%.

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?

You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds. You were also told that the probability that the time lapsed between two consecutive trades to fall between 16 to 17 seconds was 13%. The probability that the time lapsed between two consecutive trades would fall below 13 seconds was 7%. The probability is 20% that the time lapsed will be shorter how many seconds?

A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of waiting time was found to be a random variable best approximated by an exponential distribution with a mean length of waiting time equal to 2.8 minutes (i.e. the mean number of calls answered in a minute is 1/2.8). What is the probability that a randomly selected caller is placed on hold fewer than 7 minutes?

The interval between patients arriving at an outpatient clinic follows an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival to be more than 1 hour?

The amount of tea leaves in a can from a particular production line is normally distributed with μ = 110 grams and σ = 25 grams. What is the probability that a randomly selected can will contain less than 100 grams of tea leaves?

Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.89 and -1.03 is ________.

TABLE 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 $2,000. At night he works 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 Table 6-2, for a given month, what is the probability that John's income as a waiter is less than $1300?

TABLE 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 $2,000. At night he works 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 Table 6-2, the probability is 0.95 that John's commission from the jewelry store is at least how much in a given month?

TABLE 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 $2,000. At night he works 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 Table 6-2, for a given month, what is the probability that John's income as a waiter is between $1,200 and $1,600?