Question 1

Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes.What is the average number of arrivals per minute?

Accepted Answer

The answer of Patients arriving at an outpatient clinic follow...

Question 2

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,John's income as a waiter will be between what two values symmetrically distributed around the population mean 80% of the time?

Accepted Answer

The answer of Instruction 6.2
John has two jobs. For daytime...

Question 3

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 between 25 and 30 computers that require repair on a given day using a normal approximation.

Accepted Answer

The answer of Instruction 6.8
A company has 125 personal computers....

Question 4

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 less than 15 minutes?

Accepted Answer

The answer of Patients arriving at an outpatient clinic follow...

Question 5

One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because the normal distribution is used for a discrete random variable while the binomial distribution is used for a continuous random variable.

Accepted Answer

A) True 
 B)False

Question 6

Instruction 6.6
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 Instruction 6.6,what is the probability that the time interval between two consecutive defective light bulbs will be less than 10 minutes?

Accepted Answer

The answer of Instruction 6.6
Suppose the time interval between two...

Question 7

To determine the probability of getting between 2 and 4 successes in a binomial distribution,you will find the area under the normal curve between X = 1.5 and 4.5.

Accepted Answer

A) True 
 B)False

Question 8

Instruction 6.7
The interval between consecutive hits at a website is assumed to follow an exponential distribution with an average of 40 hits per minute.
-Referring to Instruction 6.7,what is the probability that the next hit at the web site will occur between the next 1.2 and 1.5 seconds after just being hit by a visitor?

Accepted Answer

The answer of Instruction 6.7
The interval between consecutive hits at...

Question 9

Theoretically,the mean,median and mode are all equal for a normal distribution.

Accepted Answer

A) True 
 B)False

Question 10

Instruction 6.7
The interval between consecutive hits at a website is assumed to follow an exponential distribution with an average of 40 hits per minute.
-Referring to Instruction 6.7,what is the probability that the next hit at the website will occur no sooner than 5 seconds after just being hit by a visitor?

Accepted Answer

The answer of Instruction 6.7
The interval between consecutive hits at...

Question 11

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

Accepted Answer

The answer of Instruction 6.2
John has two jobs. For daytime...

Question 12

Scientists in Indonesia are trying to find a cure for a deadly disease that is attacking the rain forests on Borneo and Sumatra.One of the variables that the scientists have been measuring involves the diameter of the trunk of the trees that have been affected by the disease.Scientists have calculated that the average diameter of the diseased trees is 42 centimetres.They also know that approximately 95% of the diameters fall between 32 and 52 centimetres and almost all of the diseased trees have diameters between 27 and 57 centimetres.When modelling the diameters of diseased trees,which distribution should the scientists use?

Accepted Answer

A)  Exponential distribution. 
B)  Normal distribution. 
C)  Uniform distribution. 
D)  Binomial distribution.

Question 13

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1.5 patients per hour.What is the probability that a randomly chosen arrival to be between 10 and 15 minutes?

Accepted Answer

The answer of Patients arriving at an outpatient clinic follow...

Question 14

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 commission from the jewellery store is more than $9,500?

Accepted Answer

The answer of Instruction 6.2
John has two jobs. For daytime...

Question 15

To determine the probability of getting fewer than 3 successes in a binomial distribution,you will find the area under the normal curve for X = 3.5 and below.

Accepted Answer

A) True 
 B)False

Question 16

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 more than $900?

Accepted Answer

The answer of Instruction 6.2
John has two jobs. For daytime...

Question 17

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 more than 25 but less than 30 computers that require repair on a given day using a normal approximation.

Accepted Answer

The answer of Instruction 6.8
A company has 125 personal computers....

Question 18

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,70% of the products would be assembled within _________ minutes.

Accepted Answer

The answer of The amount of time necessary for assembly...

Question 19

If a particular batch of data is approximately normally distributed,we would find that approximately

Accepted Answer

A)  19 of every 20 observations would fall between ± 2 standard deviations around the mean. 
B)  4 of every 5 observations would fall between ± 1.28 standard deviations around the mean. 
C)  2 of every 3 observations would fall between ± 1 standard deviation around the mean. 
D)  All the above.

Question 20

Instruction 6.6
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 Instruction 6.6,what is the probability that the time interval between two consecutive defective light bulbs will be at least 50 minutes?

Accepted Answer

The answer of Instruction 6.6
Suppose the time interval between two...

Exam 6: The Normal Distribution and Other Continuous Distributions

Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes.What is the average number of arrivals per minute?

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 less than 15 minutes?

One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because the normal distribution is used for a discrete random variable while the binomial distribution is used for a continuous random variable.

To determine the probability of getting between 2 and 4 successes in a binomial distribution,you will find the area under the normal curve between X = 1.5 and 4.5.

Theoretically,the mean,median and mode are all equal for a normal distribution.

Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1.5 patients per hour.What is the probability that a randomly chosen arrival to be between 10 and 15 minutes?

To determine the probability of getting fewer than 3 successes in a binomial distribution,you will find the area under the normal curve for X = 3.5 and below.

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,70% of the products would be assembled within _________ minutes.

If a particular batch of data is approximately normally distributed,we would find that approximately