Question 1

A manufacturer of personal computers tests competing brands and finds that the amount of energy they require is normally distributed with a mean of 285 kwh and a standard deviation of 9.1 kwh.If the lowest 25 percent and the highest 30 percent are not included in a second round of tests,what are the upper and lower limits for the energy amounts of the remaining computers?
A)[269.76 300.24]
B)[278.86 289.78]
C)[280.22 289.78]
D)[280.22 300.24]

Accepted Answer

To find the lower limit for the energy amounts of the remaining computers, we need to find the energy amount below which 25% of the computers lie. Using a standard normal distribution table or calculator, we find the z-score corresponding to the 25th percentile to be -0.675. We can convert this z-score to an actual energy amount using the formula:

X = μ + zσ

X = 285 + (-0.675)(9.1) = 278.86

Therefore, the lower limit is 278.86 kwh.

To find the upper limit, we need to find the energy amount below which 70% of the computers lie (since we excluded the lowest 25% and the highest 30%). Using a standard normal distribution table or calculator, we find the z-score corresponding to the 70th percentile to be 0.524. We can convert this z-score to an actual energy amount using the formula:

X = μ + zσ

X = 285 + (0.524)(9.1) = 289.78

Therefore, the upper limit is 289.78 kwh.

The only choice that has both the correct lower and upper limits is B, [278.86 289.78].

Question 2

While conducting experiments,a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.What is the probability that a randomly selected depth is less than 3.60 m?
A).72
B).80
C).46
D).32

Accepted Answer

The range of the uniformly distributed collection of water depths is from 2.00 m to 7.00 m, which means that the total range is 7.00 - 2.00 = 5.00 m. 
The probability of selecting a depth less than 3.60 m can be represented by the proportion of the range that is below 3.60 m. 
This proportion can be calculated as (3.60 - 2.00) / (7.00 - 2.00) = 1.60 / 5.00 = 0.32. 
Therefore, the probability of selecting a depth less than 3.60 m is 0.32 or 32%. The correct answer is D).

Question 3

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.What is the probability that a randomly selected individual will have a waiting time of at least 10 minutes?
A).9938
B).8944
C).1056
D).0062

Accepted Answer

P(x ≥ 10) = p= p(z ≥ −2.5) = 1.0 − .0062 = .9938

Question 4

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent more than $16.00 on sugar.
A)1.00
B)0.00
C)0.50
D)0.98

Accepted Answer

It is impossible for a household to spend more than $16.00 on sugar if the mean is $8.22 and the standard deviation is $1.10. The highest possible expenditure would be $8.22 + 3($1.10) = $11.52, which is still significantly lower than $16.00. Therefore, the probability is 0.

Question 5

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.99 percent of the households spent less than what amount?
A)$5.66
B)$10.78
C)$6.81
D)$9.63

Accepted Answer

The answer of During the past six months,73.2 percent of...

Question 6

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.What is the probability that a randomly selected individual will have a waiting time of no more than 22 minutes?
A).1587
B).9977
C).8413
D).0023

Accepted Answer

P(x ≤ 22) = p= p(z < −1) = .1587

Question 7

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.What is the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes?
A)1.00
B).9699
C).5000
D).9398

Accepted Answer

Using the normal distribution formula, we can find the z-scores for 15 and 45 minutes:
z1 = (15-30)/8 = -1.875
z2 = (45-30)/8 = 1.875
Then, using a standard normal distribution table, we can find the area between these z-scores:
P(-1.875 < Z < 1.875) = 0.9398
Therefore, the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes is 0.9398, or approximately 0.94.

Question 8

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.Complete the following statement: Only 20 percent of the individuals wait less than _____ minutes.
A)36.72
B)23.28
C)34.63
D)25.37

Accepted Answer

z_.20 = −.84 = (x − 30)/8 = −6.72 + 30 x = 23.28

Question 9

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.Only 5 percent of the students taking the test scored higher than what grade?
A)70.04
B)67.96
C)55.48
D)82.52

Accepted Answer

Using a standard normal distribution table or calculator, we can find the z-score for the top 5% of scores: 
z = 1.645 
We can then use the formula z = (x - μ) / σ to solve for x, the score we're looking for: 
1.645 = (x - 69) / 9 
x - 69 = 14.805 
x = 83.805 
Rounding to two decimal places, we get 82.52, which corresponds to choice D.

Question 10

Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distributed between 40 minutes and 65 minutes.What is the probability that a randomly selected cable repair visit will take at least 50 minutes?
A).77
B).40
C).60
D).23

Accepted Answer

p(x > 50) === .6

Question 11

While conducting experiments,a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.What is the probability that a randomly selected depth is between 2.25 m and 5.00 m?
A).79
B).45
C).55
D).50

Accepted Answer

The range of depths is 7.00 m - 2.00 m = 5.00 m.
The depth between 2.25 m and 5.00 m is a range of 5.00 m - 2.25 m = 2.75 m within the total range. 
Therefore, the probability of selecting a depth between 2.25 m and 5.00 m is 2.75 m / 5.00 m = 0.55 or 55%.

Question 12

While conducting experiments,a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.What is the standard deviation of the water depth?
A)1.4
B)2.6
C)2.5
D)4.5

Accepted Answer

== 1.44

Question 13

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.80 percent of the households spent more than what amount?
A)$7.30
B)$7.38
C)$9.06
D)$9.14

Accepted Answer

The answer of During the past six months,73.2 percent of...

Question 14

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes.Suppose that in an effort to provide better service to the public,the director of the local office is permitted to provide discounts to those individuals whose waiting time exceeds a predetermined time.The director decides that 15 percent of the customers should receive this discount.What are the numbers of minutes they need to wait to receive the discount?
A)34.48
B)21.68
C)38.32
D)25.52

Accepted Answer

First, we need to find the z-score associated with the 85th percentile (the other 15% will receive the discount). We can use a Z-table or a calculator to find that the z-score is approximately 1.04.

Then, we can use the formula for a standard normal distribution to find the corresponding value in minutes:
X = μ + (Z * σ)
X = 30 + (1.04 * 8)
X = 38.32

Therefore, customers who wait for more than 38.32 minutes will receive the discount. The closest answer choice is C, 38.32.

Question 15

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What proportion of students scored between 81 and 89?
A).0668
B).0132
C).0786
D).0529

Accepted Answer

We need to find the proportion of students between z-scores of (81-69)/9 = 1.33 and (89-69)/9 = 2.22. Using a standard normal distribution table or calculator, we can find the area between these two z-scores to be 0.0786. Therefore, approximately 7.86% of students scored between 81 and 89.

Question 16

Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distributed between 40 minutes and 65 minutes.What is the probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean?
A)0.75
B)1.00
C)0.58
D)0.86

Accepted Answer

For a uniform distribution, the probability of a value falling within 2 standard deviations of the mean is always 1.00, as the entire range of the distribution is within 2 standard deviations.

Question 17

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.What proportion of the households spent between $5.00 and $9.00 on sugar?
A).7611
B).7628
C).0017
D).7594

Accepted Answer

The answer of During the past six months,73.2 percent of...

Question 18

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What is the probability of getting a grade of 91 or less on this exam?
A)1.00
B)0.50
C)0.0073
D).9927

Accepted Answer

To find the probability of getting a grade of 91 or less, we need to find the area under the normal curve to the left of 91. We can standardize this value using the formula:
z = (x - μ) / σ
where x is the value we're interested in, μ is the mean, and σ is the standard deviation.
Substituting the values given in the problem statement, we get:
z = (91 - 69) / 9
z = 2.44
Using a standard normal distribution table or calculator, we can find that the probability of a z-score less than or equal to 2.44 is 0.9927. Therefore, the probability of getting a grade of 91 or less on this exam is 0.9927.

Question 19

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What percentage of students scored between 65 and 89?
A)65.68%
B)100%
C)98.68%
D)33.00%

Accepted Answer

The answer of A set of final examination grades in...

Question 20

While conducting experiments,a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.What is the expected value of the water depth?
A)1.4
B)2.6
C)2.5
D)4.5

Accepted Answer

== 4.5

Quiz 6: Continuous Random Variables

While conducting experiments,a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.What is the probability that a randomly selected depth is less than 3.60 m?

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.Find the probability that a household spent more than $16.00 on sugar.

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.99 percent of the households spent less than what amount?

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.Only 5 percent of the students taking the test scored higher than what grade?

Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distributed between 40 minutes and 65 minutes.What is the probability that a randomly selected cable repair visit will take at least 50 minutes?

While conducting experiments,a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.What is the probability that a randomly selected depth is between 2.25 m and 5.00 m?

While conducting experiments,a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.What is the standard deviation of the water depth?

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.80 percent of the households spent more than what amount?

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What proportion of students scored between 81 and 89?

Suppose that the times required for a cable company to fix cable problems in the homes of its customers are uniformly distributed between 40 minutes and 65 minutes.What is the probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean?

During the past six months,73.2 percent of US households purchased sugar.Assume that these expenditures are approximately normally distributed with a mean of $8.22 and a standard deviation of $1.10.What proportion of the households spent between $5.00 and $9.00 on sugar?

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What is the probability of getting a grade of 91 or less on this exam?

A set of final examination grades in a calculus course was found to be normally distributed with a mean of 69 and a standard deviation of 9.What percentage of students scored between 65 and 89?

While conducting experiments,a marine biologist selects water depths from a uniformly distributed collection that vary between 2.00 m and 7.00 m.What is the expected value of the water depth?