Question 1

Suppose the daily change in price of a stock is normally distributed with mean = .20 and standard deviation = .30.What price change is associated with the 25^th percentile? A).1925 B).2075 C).401 D)-.001

Accepted Answer

The answer of Suppose the daily change in price of...

Question 2

The weight of a product is normally distributed with a mean of four 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?
A).8413
B).9772
C).1587
D).0228

Accepted Answer

We need to find the probability that the weight of a randomly selected unit weighs no more than 3.5 ounces. 
First, we need to find the standard deviation by taking the square root of the variance: 
SD = sqrt(0.25) = 0.5 
Then, we can standardize the value of 3.5 using the z-score formula: 
z = (x - mu) / SD 
z = (3.5 - 4) / 0.5 
z = -0.5 
Using a standard normal distribution table or calculator, we can find the probability that z is less than or equal to -0.5, which is 0.1587. 
Therefore, the probability that a randomly selected unit weighs no more than 3.5 ounces is 0.1587 or approximately 15.87%. 
The answer is choice C.

Question 3

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?
A).4332
B).4878
C).0546
D).9210

Accepted Answer

The answer of The population of lengths of aluminum-coated steel...

Question 4

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X > 13).
A).0668
B).9544
C).0456
D).9332

Accepted Answer

P(X > 13) = 1 - P(X <= 13) = 1 - 0.9332 = 0.0668

Question 5

The weight of a product is normally distributed with a mean of four 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?
A).3085
B).6915
C).1587
D).8413

Accepted Answer

The answer of The weight of a product is normally...

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 0.2 inches. What is the probability that a sheet selected at random will be less than 29.75 inches long?
A).8944
B).1056
C).9332
D).0668

Accepted Answer

Using the z-score formula: (29.75-30.05)/0.2 = -1.5
Looking up the area to the left of z = -1.5 in a standard normal table or using a calculator, we find the probability to be 0.0668, or 6.68%. Therefore, the answer is D.

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% of the products to weigh more than 4.8 ounces?
A)3.52
B)3.64
C)5.44
D)4.16

Accepted Answer

To ensure that no more than 10% of the products weigh more than 4.8 ounces, we need to find the average weight such that 4.8 is at the 90th percentile of the normal distribution. Using the z-score for the 90th percentile (approximately 1.28) and the given standard deviation of 0.5, we can calculate the mean: Mean = 4.8 - (1.28 * 0.5) = 4.16.

Question 8

Consider a normal population with a mean of 10 and a variance of 4.Find P(X $\ge$ 10).
A)1.00
B)0.00
C)0.50
D)-0.50

Accepted Answer

The answer of Consider a normal population with a mean...

Question 9

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X < 12).
A).1587
B).9544
C).0456
D).8413

Accepted Answer

The Z-score for X = 12 is (12-10)/2 = 1. The probability P(X < 12) corresponds to P(Z < 1) in the standard normal distribution, which is 0.8413.

Question 10

Consider a normal population with a mean of 10 and a variance of 4.Find P(X > 7).
A).0668
B).9332
C).8413
D).1587

Accepted Answer

We need to find the probability of getting a value greater than 7 from a normal distribution with mean 10 and variance 4. 
First, we need to find the standard deviation (σ) by taking the square root of the variance: 
σ = √4 = 2 
Next, we need to standardize the value of 7 to a z-score using the formula: 
z = (X - μ) / σ 
z = (7 - 10) / 2 
z = -1.5 
Using a standard normal distribution table or calculator, we can find that the probability of getting a value greater than -1.5 is 0.9332. 
However, we need the probability of getting a value greater than 7, so we need to subtract this value from 1: 
P(X > 7) = 1 - P(X ≤ 7) = 1 - 0.9332 = 0.0668 
Therefore, the answer is B.

Question 11

The weight of a product is normally distributed with a mean of four ounces and a variance of .25 "squared ounces." The company wants to classify the unit as a scrap in a maximum of 1% of the units if the weight is below a desired value.Determine the desired weight such that no more than 1% of the units are below it.
A)3.360
B)3.680
C)2.835
D)3.418

Accepted Answer

The answer of The weight of a product is normally...

Question 12

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X = 10).
A)1.00
B)0.99
C)0.00
D)0.01

Accepted Answer

In a continuous probability distribution like the normal distribution, the probability of any single exact value is 0.

Question 13

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% of the products to weigh more than 5.1 ounces?
A)4.278
B)4.409
C)3.455
D)5.922

Accepted Answer

To ensure no more than 5% of products weigh more than 5.1 ounces, the average weight should be set such that 5.1 is at the 95th percentile of the normal distribution. Using the z-score for 95% (approximately 1.645) and the given standard deviation (0.5), the average weight can be calculated as: 5.1 - (1.645 * 0.5) = 4.278.

Question 14

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long?
A).9987
B).1574
C).1587
D).8413

Accepted Answer

The answer of The population of lengths of aluminum-coated steel...

Question 15

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.
A).5675
B).0170
C).5714
D).4325

Accepted Answer

The answer of If x is a binomial random variable...

Question 16

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random will be less than 31 inches long?
A)0.00
B)1.00
C).8289
D).5987

Accepted Answer

The answer of The population of lengths of aluminum-coated steel...

Question 17

Consider a normal population with a mean of 10 and a variance of 4.Find P(X < 6).
A).0228
B).1587
C).8413
D).9772

Accepted Answer

To find P(X<6), we first need to standardize the value of 6 using the formula:

Z = (X - µ) / σ

Where µ is the mean and σ is the standard deviation. Since the variance is given, we can find the standard deviation by taking the square root of the variance: σ = sqrt(4) = 2.

So, Z = (6 - 10) / 2 = -2

Using a standard normal distribution table, we can find the probability of Z being less than -2. This is equivalent to the area to the left of the Z-score of -2. The value we get from the table is 0.0228.

Therefore, P(X<6) = P(Z<-2) = 0.0228. The answer is A).

Question 18

If the mileage per gallon for a car is normally distributed,32 mpg has a z-score of 1.2,and 24 mpg has a z-score of -.4,what is the mean mpg of the distribution?
A)28
B)26
C)30
D)38

Accepted Answer

μ_x + Zσ_x = X μ_x + 1.2σ_x = 32 (1) μ_x + 4σ_x = 24 (2) μ_x + 32 − 1.2σ_x, substituting μ_xin (2); 32 − 1.2σ_x− .4σ = 24 1.6σ = 8 σ = 5 μ_x = 32 − 1.2(5) = 26

Question 19

The weight of a product is normally distributed with a mean of four 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?
A).0228
B).9772
C)0.000
D)1.00

Accepted Answer

First, we need to calculate the standard deviation which is the square root of the variance:
$\sqrt{0.25}=0.5$
Next, we need to standardize the value of 5 ounces using the formula:
$z= \frac{x-\mu}{\sigma}$
Where $x=5$, $\mu=4$, and $\sigma=0.5$
$z= \frac{5-4}{0.5}=2$
Using a standard normal distribution table, we can find that the probability of z being greater than 2 is approximately 0.0228. Therefore, the probability of a randomly selected unit from the batch weighing more than 5 ounces is 0.0228 or 2.28%. The best choice is (A).

Question 20

Consider a normal population with a mean of 10 and a variance of 4.Find P(X > 18).
A)1.00
B)0.00
C).9772
D).0228

Accepted Answer

Since the distribution is normal, we can use the z-score formula: Z = (X - μ) / σ.

Plugging in the values, we get: 
Z = (18 - 10) / √4 = 4

Looking up the z-score of 4 in a standard normal distribution table, we find that the probability of getting a value greater than that is basically 0 (to the nearest hundredth), which corresponds to choice B.

Quiz 6: Continuous Random Variables

Suppose the daily change in price of a stock is normally distributed with mean = .20 and standard deviation = .30.What price change is associated with the 25^th percentile?

The weight of a product is normally distributed with a mean of four 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?

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X > 13) .

The weight of a product is normally distributed with a mean of four 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?

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random will be less than 29.75 inches long?

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% of the products to weigh more than 4.8 ounces?

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

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X < 12) .

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

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X = 10) .

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% of the products to weigh more than 5.1 ounces?

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long?

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.

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random will be less than 31 inches long?

Consider a normal population with a mean of 10 and a variance of 4.Find P(X < 6) .

If the mileage per gallon for a car is normally distributed,32 mpg has a z-score of 1.2,and 24 mpg has a z-score of -.4,what is the mean mpg of the distribution?

The weight of a product is normally distributed with a mean of four 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?

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

Quiz 6: Continuous Random Variables

Suppose the daily change in price of a stock is normally distributed with mean = .20 and standard deviation = .30.What price change is associated with the 25th percentile?

The weight of a product is normally distributed with a mean of four 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?

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 29.75 and 30.5 inches long?

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X > 13) .

The weight of a product is normally distributed with a mean of four 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?

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random will be less than 29.75 inches long?

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% of the products to weigh more than 4.8 ounces?

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

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X < 12) .

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

Consider a normal population with a mean of 10 and a standard deviation 2. Find P(X = 10) .

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% of the products to weigh more than 5.1 ounces?

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long?

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.

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random will be less than 31 inches long?

Consider a normal population with a mean of 10 and a variance of 4.Find P(X < 6) .

If the mileage per gallon for a car is normally distributed,32 mpg has a z-score of 1.2,and 24 mpg has a z-score of -.4,what is the mean mpg of the distribution?

The weight of a product is normally distributed with a mean of four 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?

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

Suppose the daily change in price of a stock is normally distributed with mean = .20 and standard deviation = .30.What price change is associated with the 25^th percentile?

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