Exam 6: The Normal Distribution and Other Continuous Distributions

The owner of a fish market determined that the average weight for a rainbow trout is 3.2 kilograms with a standard deviation of 0.8 kilogram.Assuming the weights of rainbow trout are normally distributed,the probability that a randomly selected rainbow trout will weigh between 3 and 5 kilograms is _________

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?

Instruction 6.2 John has two jobs. For daytime work at a jewellery 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 Instruction 6.2,for a given month,what is the probability that John's income as a waiter is between $1,200 and $1,600?

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. -So,60% of the products would be assembled within _________ and _________ minutes (symmetrically distributed about the mean).

As a general rule,one can use the normal distribution to approximate a binomial distribution whenever np is at least 5.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour.What is the probability that a randomly chosen arrival to be between 5 minutes and 15 minutes?

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

One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because a random variable having a binomial distribution can have only a specified value while a random variable having a normal distribution can take on any values within an interval around that specified value.

Instruction 6.1 The number of column centimetres of classified advertisements appearing on Mondays in a certain daily newspaper is normally distributed with population mean 320 and population standard deviation 20 cm. -Referring to Instruction 6.1,for a randomly chosen Monday the probability is 0.1 that there will be less than how many column centimetres of classified advertisements?

Let X represent the amount of time it takes a student to park in the car park at the university.If we know that the distribution of parking times can be modelled using an exponential distribution with a mean of 4 minutes,find the probability that it will take a randomly selected student between 2 and 12 minutes to park in the car park.

For some positive value of Z,the value of the cumulative standardised normal distribution is 0.8340.The value of Z is

Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes.What is the probability that a randomly chosen arrival to be less than 15 minutes?

Instruction 6.8 A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.15. -Referring to Instruction 6.8,which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than two computers will require repair on a given day?

The amount of pyridoxine (in grams)in a multiple vitamin is normally distributed with $\mu$ = 110 grams and $\sigma$ 25 grams. -What is the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine?

The value of the cumulative standardised normal distribution at Z is 0.6255.The value of Z 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 between 15 and 21 minutes.

If we know that the length of time it takes a university student to find a parking space in the university car park follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute,find the point in the distribution in which 75.8% of the university students exceed when trying to find a parking space in the car park.

Instruction 6.8 A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.15. -Referring to Instruction 6.8 and assuming that the number of computers that requires repair on a given day follows a binomial distribution,compute the probability that there will be less than 8 computers that require repair on a given day using a normal approximation.

Which of the following can be used to model the distribution of the values for a continuous random variable?

Explain why the exponential distribution is frequently used in waiting line (or queuing)theory,and provide three examples of what the exponential random variable can be used to model.