Exam 6: The Normal Distribution

The "middle spread," that is, the middle 50% of the normal distribution, is equal to one standard deviation.

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.

The owner of a fish market determined that the average 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 less than 2.2 pounds 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 less than 12 minutes.

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%. What is the probability that the time lapsed between two consecutive trades will be between 14 and 15 seconds?

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 at least $1,400?

The probability that a standard normal random variable, Z, is less than 5.0 is approximately 0.

A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces.

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 ________.

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?

In its standardized form, the normal distribution

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 over 125 inches in length?

The owner of a fish market determined that the average 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 ________.

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, John's income as a waiter will be between what two values symmetrically distributed around the population mean 90% of the time?

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 0.10 that John's commission from the jewelry store is more than how much in a given month?

The probability that a standard normal random variable, Z, falls between -2.00 and -0.44 is 0.6472.

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 $700 and $1,600?

The probability that a particular brand of smoke alarm will function properly and sound an alarm in the presence of smoke is 0.8. A batch of 100,000 such alarms was produced by independent production lines. Which of the following distributions would you use to figure out the probability that at least 90,000 of them will function properly in case of a fire?

Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.33 and 2.33 is ________.

TABLE 6-4 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 6-4, find the probability that the annual return of a random year will be less than 11.5%.