Question 1

Consider a normal population with a mean of 10 and a variance of 4. Find P(X > 18).

Accepted Answer

Z == 4
P(Z > 4) ≅ 0

Question 2

The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. What is the probability someone will spend at least one hour reading online national news reports?

Accepted Answer

P (x ≥ 60)
PPP (z > .69)
1.00 − .7541 = .2459

Question 3

The weight of a product is normally distributed with a mean of 4 ounces and a variance of .25 squared ounces. The company wants to classify the unit as a scrap in a maximum of 1 percent of the units if the weight is below a desired value. Determine the desired weight such that no more than 1 percent of the units are below it.

Accepted Answer

X = μ − Z σ = 4 − (2.33)= 2.835 ounces

Question 4

The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. For the 10 percent who spend the most time reading national news online, how much time do they spend?

Accepted Answer

= 1.28125 =x = 20.50 + 49 = 69.50

Question 5

The weight of a product is normally distributed with a mean of 5 ounces. A randomly selected unit of this product weighs 7.1 ounces. The probability of a unit weighing more than 7.1 ounces is .0014. The production supervisor has lost files containing various pieces of information regarding this process, including the standard deviation. Determine the value of the standard deviation for this process.

Accepted Answer

σ ==== .7

Question 6

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of .2 inches. What is the probability that a sheet selected at random will be less than 31 inches long?

Accepted Answer

P(z ≤ [31.00 − 30.05]/.2) = P(z ≤ 4.75) = approx. 1

Question 7

The weight of a product is normally distributed with a standard deviation of .5 ounces. What should the average weight be if the production manager wants no more than 10 percent of the products to weigh more than 4.8 ounces?

Accepted Answer

P(Z > 1.28) = .10= X − Z= 4.8 − (1.28) (.5) = 4.16

Question 8

The weight of a product is normally distributed with a mean of 4 ounces and a variance of .25 squared ounces. What is the probability that a randomly selected unit from a recently manufactured batch weighs no more than 3.5 ounces?

Accepted Answer

z = (3.5 − 4)/= −1
P(z < −1) = .1587

Question 9

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of .2 inches. What is the probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long?

Accepted Answer

P([30.25 − 30.05]/.2 ≤ z ≤ (30.65 − 30.05)/.2) = P(1 ≤ z ≤ 3.0) = .9987 − .8413 = .1574

Question 10

The weight of a product is normally distributed with a mean of 4 ounces and a variance of .25 squared ounces. What is the probability that a randomly selected unit from a recently manufactured batch weighs more than 5 ounces?

Accepted Answer

z = (5 − 4)/= 2
P(z > 2) = 1 − .9772

Question 11

The weight of a product is normally distributed with a standard deviation of .5 ounces. What should the average weight be if the production manager wants no more than 5 percent of the products to weigh more than 5.1 ounces?

Accepted Answer

P(Z > 1.645) = 0.5= X − Z= 5.1 − (1.645) (.5) = 4.278

Question 12

If x is a binomial random variable where n = 100 and p = .1, find the probability that x is less than or equal to 10, using the normal approximation to the binomial.

Accepted Answer

z = (10.5 − 10)/√[(.1)(.9)(100)] = .5/3 = .017 approx.
P(X ≤ 10) = .5662

Question 13

The weight of a product is normally distributed with a mean of 4 ounces and a variance of .25 squared ounces. What is the probability that a randomly selected unit from a recently manufactured batch weighs more than 3.75 ounces?

Accepted Answer

z = (3.75 − 4)/= −.5
P(z > −.5) = 1 − .3085 = .6915

Question 14

Consider a normal population with a mean of 10 and a variance of 4. Find P(X > 7).

Accepted Answer

z = (7 − 10)/= −1.5
P(X > 7) = 1 − .0668 = .9332

Question 15

The average time an individual reads online national news reports is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. What is the probability someone will spend no more than 30 minutes reading online national news reports?

Accepted Answer

P (x < 30)
PP (z < −1.1875) = .1175

Question 16

At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water that is discharged back into the ocean. The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25° C. What is the probability that the temperature increase will be between 20° and 22° C?

Accepted Answer

P (20 ≤ x ≤ 22) === .13

Question 17

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of .2 inches. What is the probability that a sheet selected at random will be less than 29.75 inches long?

Accepted Answer

P(z ≤ [29.75 − 30.05]/.2) = P(Z ≤ −1.5) = .0668

Question 18

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of .2 inches. What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?

Accepted Answer

P(−1.5 ≤ z ≤ 0) = .5 − .0668 = .4332
P(0 ≤ z ≤ 2.25) = .9878 − .5000 = .4878
.4332 + .4878 = .921

Question 19

At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water that is discharged back into the ocean. The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25° C. What is the probability that the temperature increase will be less than 20° C?

Accepted Answer

P (10 ≤ x ≤ 20) === .67

Question 20

Consider a normal population with a mean of 10 and a variance of 4. Find P(X ≥ 10).

Accepted Answer

z = 0; P(z ≥ 0) = .5

Quiz 7: Continuous Random Variables

Consider a normal population with a mean of 10 and a variance of 4. Find P(X > 18) .