Question 1

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?

Accepted Answer

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

Question 2

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

Since the temperature increase has a uniform distribution from 10° to 25° C, the probability density function (PDF) is a horizontal line with height 1/15 over that interval. The probability of the temperature increase being between 20° and 22° C is the area under the PDF between those two values. This area is a rectangle with base 2 and height 1/15, so its area is (2)(1/15) = 2/15. Therefore, the probability is 2/15 ≈ 0.1333, which is closest to choice C.

Question 3

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

The normal approximation to the binomial distribution can be used here. The mean (μ) is np = 100 * 0.1 = 10 and the standard deviation (σ) is sqrt(np(1-p)) = sqrt(100 * 0.1 * 0.9) = 3. The continuity correction factor is applied, so we find P(X ≤ 10.5). Using the standard normal distribution, this corresponds to a z-score of (10.5 - 10) / 3 = 0.1667. The probability corresponding to this z-score is approximately 0.5675.

Question 4

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

Since the temperature increase has a uniform distribution over the interval from 10° to 25° C, the probability of it being less than 20° C is the proportion of the length of the interval from 10° to 20° C to the length of the whole interval 10° to 25° C:

$P(	ext{temperature increase}<20°) = \frac{20-10}{25-10}=\frac{10}{15}=\frac{2}{3}=0.67$

Therefore, the answer is B.

Question 5

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?

Accepted Answer

We need to find the probability that a randomly selected sheet will be less than 29.75 inches long. We can use the standard normal distribution to find this probability.

First, we need to standardize the value of 29.75 using the formula:

z = (x - μ) / σ

where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.

Plugging in the values, we get:

z = (29.75 - 30.05) / 0.2 = -1.5

Now, we need to find the probability that z is less than -1.5. We can use a standard normal distribution table or calculator to find this probability.

From the table, we find that the probability of z being less than -1.5 is approximately 0.0668.

Therefore, the probability that a sheet selected at random will be less than 29.75 inches long is approximately 0.0668, or 6.68%.

The closest answer choice is D) 0.0668, which is the correct answer.

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 from the population is between 29.75 and 30.5 inches long?

Accepted Answer

The probability can be found by calculating the z-scores for 29.75 and 30.5, then finding the area between these z-scores in the standard normal distribution. The z-score for 29.75 is (29.75 - 30.05) / 0.2 = -1.5, and for 30.5 is (30.5 - 30.05) / 0.2 = 2.25. The area between these z-scores corresponds to a probability of approximately 0.9210.

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 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 8

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 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

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

Question 9

The weight of a product is normally distributed with a mean 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

The z-score corresponding to a probability of 0.0014 in the upper tail of a standard normal distribution is approximately 3.00. Using the z-score formula $ z = \frac{X - \mu}{\sigma} $, where $ X = 7.1 $, $ \mu = 5 $, and $ z = 3.00 $, we solve for $ \sigma $: $ 3.00 = \frac{7.1 - 5}{\sigma} $. This gives $ \sigma = \frac{2.1}{3.00} = 0.70 $.

Question 10

The average time a subscriber spends reading the local newspaper 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 the paper,how much time do they spend?

Accepted Answer

To find the time spent by the top 10% of readers, we need to find the z-score that corresponds to the 90th percentile of a standard normal distribution, which is approximately 1.28. Using the formula: X = μ + zσ, where μ = 49, σ = 16, and z = 1.28, we calculate X = 49 + (1.28)(16) = 69.48 minutes.

Question 11

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

Accepted Answer

To find P(X ≥ 10), we need to find the area to the right of 10 on a standard normal distribution curve. However, we don't have a standard normal distribution in this case, we have a normal population with mean 10 and variance 4.

We can standardize the distribution by using the z-score formula:

z = (x - μ) / σ

where x is the value we're interested in (in this case, 10), μ is the mean of the population (10), and σ is the standard deviation of the population (sqrt(4) = 2).

Plugging in the values, we get:

z = (10 - 10) / 2 = 0

This tells us that 10 is exactly at the mean of the population, so the area to the right of 10 (i.e. P(X ≥ 10)) is exactly 0.5.

Therefore, the answer is C.

Question 12

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?

Accepted Answer

To solve this problem, we need to standardize the weight using z-score formula:
$z=\frac{x-\mu}{\sigma}$
where:
$\mu=4$ (mean)
$\sigma=\sqrt{0.25}=0.5$ (standard deviation)
$x=3.5$ (weight we want to find the probability for)

$z=\frac{3.5-4}{0.5}=-1$
Now we can find the probability of a standard normal variable being less than or equal to -1 using normal distribution table, which gives us:
$P(Z\leq -1)=0.1587$
Therefore, the probability that a randomly selected unit from the batch weighs no more than 3.5 ounces is 0.1587 or 15.87%.
The correct answer is (C).

Question 13

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

Accepted Answer

To find P(X > 7), we first need to standardize the variable using the formula z = (X - μ) / σ, where X is the value we're interested in (in this case, X = 7), μ is the mean of the population (μ = 10), and σ is the standard deviation of the population (σ = sqrt(4) = 2).

z = (7 - 10) / 2 = -1.5

Next, we look up the probability of Z being greater than -1.5 in the standard normal distribution table, which is 0.9332. Therefore, the answer is B: 0.9332.

Question 14

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

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.

Question 15

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 16

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?

Accepted Answer

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

Question 17

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?

Accepted Answer

To solve the problem, we need to standardize the variable x (the weight of the product) using the formula:

z = (x - μ) / σ

where μ is the mean and σ is the standard deviation (not the variance).
In this case, μ = 4 ounces and σ² = 0.25 squared ounces, so σ = 0.5 ounces.
Therefore,

z = (x - 4) / 0.5

Now we want to find the probability that x > 5 ounces, which is equivalent to finding the probability that z > (5 - 4) / 0.5 = 2.

Using a standard normal distribution table or calculator, we find that P(z > 2) = 0.0228. Therefore, the answer is A) 0.0228.

Question 18

The average time a subscriber spends reading the local newspaper is 49 minutes.Assume the standard deviation is 16 minutes and that the times are normally distributed.What is the probability a subscriber will spend at least 1 hour reading the paper?

Accepted Answer

To find the probability that a subscriber spends at least 1 hour (60 minutes) reading the paper, we first calculate the z-score: z = (60 - 49) / 16 = 0.6875. Using the standard normal distribution table, the probability of a z-score less than 0.6875 is approximately 0.7549. Therefore, the probability of spending at least 60 minutes is 1 - 0.7549 = 0.2451.

Question 19

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?

Accepted Answer

Since the mean is 30.05 inches and the standard deviation is 0.2 inches, a length of 31 inches is 4.75 standard deviations above the mean. In a normal distribution, almost all values lie within 3 standard deviations of the mean, so the probability of a value being less than 31 inches is essentially 1.00.

Question 20

The average time a subscriber spends reading the local newspaper is 49 minutes.Assume the standard deviation is 16 minutes and that the times are normally distributed.What is the probability a subscriber will spend no more than 30 minutes reading the paper?

Accepted Answer

The answer of The average time a subscriber spends reading...

Quiz 6: Continuous Random Variables

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 29.75 inches long?

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?

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?

The average time a subscriber spends reading the local newspaper 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 the paper,how much time do they spend?

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

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?

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 variance of 4.Find P(X > 18) .

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?

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 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?

The average time a subscriber spends reading the local newspaper is 49 minutes.Assume the standard deviation is 16 minutes and that the times are normally distributed.What is the probability a subscriber will spend at least 1 hour reading the paper?

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?

The average time a subscriber spends reading the local newspaper is 49 minutes.Assume the standard deviation is 16 minutes and that the times are normally distributed.What is the probability a subscriber will spend no more than 30 minutes reading the paper?