Week 4 Test
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American Military University *
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Course
302
Subject
Statistics
Date
Jan 9, 2024
Type
docx
Pages
9
Uploaded by bradymcdede
Attempt Score
20 / 20 - 100 %
Overall Grade (Highest Attempt)
20 / 20 - 100 %
stion 1
1 / 1 p
Find P(Z ≤ 2.76). Round answer to 4 decimal places. Answer:
___0.9971
___
uestion 1 feedback
el,
M.S.DIST(2.76,TRUE)
on 2
1 / A dishwasher has a mean life of 11.5 years with an estimated standard deviation of 1.5 years ("Appliance life expectancy," 2013). Assume the life of a dishwasher is normally distributed.
Find the number of years that the bottom 10% of dishwasher would last. Round answer to 2 decimal places. Answer:
___9.58
___
uestion 2 feedback
el,
M.INV(0.1,11.5,1.5)
on 3
1 / Find P(Z ≥ .42). Round answer to 4 decimal places. Answer:
___.3372
___
uestion 3 feedback
el,
RM.S.DIST(0.42,TRUE)
on 4
1 / Which type of distribution does the graph illustrate?
Poisson Distribution
Right skewed Distribution
Uniform Distribution
Normal Distribution
Question 5
1 / 1
point
The cost of unleaded gasoline in the Bay Area once followed a normal distribution with a mean of $4.74 and a standard deviation of $0.16. Fifteen gas stations from the Bay area are randomly chosen. We are interested in the average cost of gasoline for the 15 gas stations. What is the approximate probability that the average price for 15 gas stations is over $4.99?
.0256
0.0000
0.0943
0.1587
Hide question 5 feedback
New SD = .16/SQRT(15) = .0413
P(x > 4.99) = 1 - P(x < 4.99) In Excel, =1-NORM.DIST(4.99,4.74,.0413,TRUE)
You might get an answer with an "E" in it. The "E"; means scientific notation. 7.1E-10 decimal answer is, .00000000071
n 6
1
Suppose that the longevity of a light bulb is exponential with a
mean lifetime of 7.6 years. Find the probability that a light bulb lasts less than three years.
0.1175
0.3261
8.5634
0.6739
0.3907
Hide question 6 feedback
P(x < 3)
In Excel,
=EXPON.DIST(3,1/7.6,TRUE)
n 7
1 Suppose that the longevity of a light bulb is exponential with a
mean lifetime of 7.6 years. Find the probability that a light bulb lasts between seven and eleven years.
0.1325
0.1859
0.1629
0.8371
0.1793
Hide question 7 feedback
P( 7 < x < 11)
P(x < 11) - P( x < 7)
In Excel,
=EXPON.DIST(11,1/7.6,TRUE)-EXPON.DIST(7,1/7.6,TRUE)
n 8
1 The commute time for people in a city has an exponential distribution with an average of 0.66 hours. What is the probability that a randomly selected person in this city will
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have a commute time between 0.55 and 1.1 hours? Answer: (round to 3 decimal places)
___0.246
___
uestion 8 feedback
x < 1.1)
.1) - P(x < .55)
el,
ON.DIST(1.1,1/0.66,TRUE)-EXPON.DIST(0.55,1/0.66,TRUE)
on 9
1 / The average lifetime of a certain new cell phone is 4.2 years. The manufacturer will replace any cell phone failing within three years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. The decay rate is:
0.2381
0.7619
0.3333
0.6667
Hide question 9 feedback
1/4.2
n 10
1 Miles per gallon of a vehicle is a random variable with a uniform distribution from 22 to 39. The probability that a random vehicle gets between 26 and 31 miles per gallon is: Answer: (Round to four decimal places)
___.2941
___
uestion 10 feedback
l goes from 22 ≤ x≤39
�
<31)=(31−26)
∗
139−22
on 11
1 / A local pizza restaurant delivery time has a uniform distribution over 10 to 75 minutes. What is the probability that
the pizza delivery time is more than 45 minutes on a given day? Answer: (Round to four decimal places.)
___0.4615
___
uestion 11 feedback
l goes from 10 < x < 75
5) = (75 - 45) *175−10
on 12
1 / The waiting time for a bus has a uniform distribution between 2 and 13 minutes. What is the probability that the waiting time for this bus is less than 4.5 minutes on a given day? Answer: (Round to four decimal places.)
___0.2273
___
uestion 12 feedback
l goes from 2 < x < 13
.5) = (4.5 - 2) *113−2
on 13
1 / The waiting time for a bus has a uniform distribution between 2 and 11 minutes. What is the 75th percentile of this distribution? _______ minutes Answer: (Round answer to two decimal places.)
___8.75
___
uestion 13 feedback
x) = .75
�
−211−2
= x- 2
2 = x
x
on 14
1 / Miles per gallon of a vehicle is a random variable with a uniform distribution from 23 to 47. The probability that a random vehicle gets between 25 and 30 miles per gallon is: Answer: (Round to four decimal places)
___0.2083
___
uestion 14 feedback
l goes from 23 < x < 47
x < 30) = (30 - 25) *147−23
on 15
1 / The waiting time for a table at a busy restaurant has a uniform
distribution between 15 and 45 minutes. What is the 90th percentile of this distribution? (Recall: The 90th percentile divides the distribution into 2 parts so that 90% of area is to the left of 90th percentile) _______ minutes Answer: (Round answer to one decimal place.)
___42
___
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uestion 15 feedback
l goes from 15 < x < 45. 90th percentile use .90
x) = .90
�
−1545−15
0 = x - 15
- 15
on 16
1 / The average amount of a beverage in randomly selected 16-
ounce beverage can is 15.85 ounces with a standard deviation
of 0.3 ounces. If a random sample of thirty-five 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 15.7 ounces of beverage? Answer: (round to 4 decimal places)
___0.0015
___
uestion 16 feedback
D =.3/SQRT(35) = 0.050709
5.7), in Excel
M.DIST(15.7,15.85,0.050709,TRUE)
on 17
1 / The final exam grade of a mathematics class has a normal distribution with mean of 81 and standard deviation of 6.6. If a
random sample of 40 students selected from this class, then what is the probability that the average final exam grade of this sample is between 77 and 82? Answer: (round to 4 decimal places)
___0.8310
___
uestion 17 feedback
D = 6.6/SQRT(40) = 1.043552
x < 82), in Excel
M.DIST(82,81,1.043552,TRUE)-NORM.DIST(77,81,1.043552,TRUE)
on 18
1 / The average amount of a beverage in randomly selected 16-
ounce beverage can is 16.18 ounces with a standard deviation
of 0.4 ounces. If a random sample of sixty-five 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 16.1 ounces of beverage? Answer: (round to 4 decimal places) ___0.0534
___
uestion 18 feedback
D = .4/SQRT(65) = 0.049614
6.1), in Excel
M.DIST(16.1,16.18,0.049614,TRUE)
on 19
1 / The MAX light rail in Portland, OR has a waiting time that is normally distributed with a mean waiting time of 4.22 minutes
with a standard deviation of 1.7 minutes. A random sample of 35 wait times was selected, what is the probability the sample
mean wait time is under 3.74 minutes? Round answer to 4 decimal places. Answer:
___0.0474
___
uestion 19 feedback
D = 1.7/SQRT(35) =0.287352
.74), using Excel
M.DIST(3.74,4.22,0.287352,TRUE)
on 20
1 / The final exam grade of a statistics class has a skewed distribution with mean of 79.8 and standard deviation of 8.2. If
a random sample of 45 students selected from this class, then
what is the probability that the average final exam grade of this sample is between 80 and 83? Answer: (round to 4 decimal places)
___0.4306
___
Hide question 20 feedback
New SD = 8.2/SQRT(45) = 1.222384
P(80 < x < 83), in Excel
=NORM.DIST(83,79.8,1.222384,TRUE)-NORM.DIST(80,79.8,1.222384,TRUE)
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