Download 1) If we know that the length of time it takes a college

1) If we know that the length of time it takes a college student to find a parking spot in
the library parking lot follows a normal distribution with a mean of 3.5 minutes and a
standard deviation of 1 minute, find the probability that a randomly selected college
student will find a parking spot in the library parking lot in less than 3 minutes.
Calculate the z-statistic: z = (3 – 3.5)/1 = -0.5
The area to the left of z = -0.5 is: 0.3085
0.2674
0.3085
0.3551
0.1915
2) If we know that the length of time it takes a college student to find a parking spot in
the library parking lot follows a normal distribution with a mean of 3.5 minutes and a
standard deviation of 1 minute, find the probability that a randomly selected college
student will take between 2 and 4.5 minutes to find a parking spot in the library parking
lot.
Calculate the z-statistic for 4.5 minutes: z = (4.5 – 3.5) / 1 = 1.0
Calculate the z-statistic for 2 minutes: z = (2 – 3.5) / 1 = -1.5
The area under the normal curve to the left of z = 1 is 0.8413
The area under the normal curve to the left of z = -1.5 is 0.0668
The area between the two z-values is 0.8413 – 0.0668 = 0.7745
0.7745
0.4938
0.2255
0.0919
3) The owner of a fish market determined that the average weight for a catfish is 3.2
pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top
2% in weight. Assuming the weights of catfish are normally distributed, at what weight
(in pounds) should the citation designation be established?
The z-value corresponding to 98% (same as the top 2%) is 2.054
Then: 2.054 = (x – 3.2)/0.8  x = 4.84
5.20 pounds
1.56 pounds
4.84 pounds
7.36 pounds
4) Suppose that prices of a certain model of new homes are normally distributed with a
mean of $150,000. Find the percentage of buyers who paid:
less than $145,200 if the standard deviation is $2400.
Calculate the z-statistic: z = (145,200 – 150,000)/2,400 = -2
The area under the normal curve to the left of z = -2 is 0.0228
47.5%
95%
2.5% (closest answer)
97.5%
5) The amount of soda a dispensing machine pours into a 12 ounce can of soda follows a
normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.
The cans only hold 12.50 ounces of soda. Every can that has more than 12.50 ounces of
soda poured into it causes a spill and the can needs to go through a special cleaning
process before it can be sold. What is the probability a randomly selected can will need to
go through this process?
Calculate the z-statistic: z = (12.50 – 12.30) / 0.20 = 1.0
The area under the normal curve to the right of z = 1.0 is 0.1587
.6587
.3413
.8413
.1587
6) A new phone system was installed last year to help reduce the expense of personal
calls that were being made by employees. Before the new system was installed, the
amount being spent on personal calls followed a normal distribution with an average of
$500 per month and a standard deviation of $50 per month. Refer to such expenses as
PCE's (personal call expenses). Using the distribution above, what is the probability that
a randomly selected month had a PCE of between $375.00 and $590.00?
Calculate the z-statistic for PCE = 590.00: z = (590 – 500)/50 = 1.8
Calculate the z-statistic for PCE = 375.00: z = (375 – 500)/50 = -2.5
The area under the normal curve to the left of z = 1.8 is 0.9641
The area under the normal curve to the left of z = -2.5 is 0.0062
The area between the two limits is 0.9641 – 0.0062 = 0.9579
.0421
.9999
.9579
.0001
7) The tread life of a particular brand of tire is a random variable best described by a
normal distribution with a mean of 60,000 miles and a standard deviation of 2000 miles.
What is the probability a particular tire of this brand will last longer than 58,000 miles?
Calculate the z-statistic: z = (58,000 – 60,000)/2000 = -1.0
The area under the normal curve to the right of z = -1.0 is 0.8413
.2266
.8413
.7266
.1587
8) A physical fitness association is including the mile run in its secondary-school fitness
test. The time for this event for boys in secondary school is known to possess a normal
distribution with a mean of 470 seconds and a standard deviation of 60 seconds. Between
what times do we expect most (approximately 95%) of the boys to run the mile?
The z-value that corresponds with an area to the left equal to 0.95 is z = 1.645
Then (x – 470) / 60 = 1.645  x = 568.7 seconds
between 371.3 and 568.736 seconds
between 0 and 568.736 seconds
between 375 and 565 seconds
between 352.4 and 587.6 seconds
9) The amount of corn chips dispensed into a 32-ounce bag by the dispensing machine
has been identified as possessing a normal distribution with a mean of 32.5 ounces and a
standard deviation of 0.2 ounce. What chip amount represents the 67th percentile for the
bag weight distribution?
The z-value that corresponds to the 67th percentile is 0.4399
Then (x – 32.5)/0.2 = 0.4399  x = 32.59
32.09 ounces
32.63 ounces
32.59 ounces
32.13 ounces
10) Suppose a brewery has a filling machine that fills 12 ounce bottles of beer. It is
known that the amount of beer poured by this filling machine follows a normal
distribution with a mean of 12.28 ounces and a standard deviation of 0.04 ounce. The
company is interested in reducing the amount of extra beer that is poured into the 12
ounce bottles. The company is seeking to identify the highest 1.5% of the fill amounts
poured by this machine. For what fill amount are they searching?
The z-value that corresponds with the top 1.5% is 2.17
Then (x – 12.28)/0.04 = 2.17  x = 12.367
12.087
12.367
11.913
12.193
11) The length of time it takes college students to find a parking spot in the library
parking lot follows a normal distribution with a mean of 3.5 minutes and a standard
deviation of 1 minute. Find the cut-off time which 75.8% of the college students exceed
when trying to find a parking spot in the library parking lot.
The z-value that corresponds with an area to the right of 0..758 is -0.6999
Then (x – 3.5) / 1 = -0.6999  x = 2.8 (which doesn’t match any of the answers)
If we interpret “exceed” to mean that 75.8% of the students find a spot in LESS
then the cut-off time then the result is:
z = +0.6999
(x – 3.5)/1 = 0.6999  x = 4.2
This is a poorly worded question, so you’ll have to use your own judgment about
which answer to use here.
3.8
4.3
4.0
4.2
12) The amount of soda a dispensing machine pours into a 12 ounce can of soda follows
a normal distribution with a standard deviation of 0.24 ounce. Every can that has more
than 12.60 ounces of soda poured into it causes a spill and the can needs to go through a
special cleaning process before it can be sold. What is the mean amount of soda the
machine should dispense if the company wants to limit the percentage that need to be
cleaned because of spillage to 3%?
The z-value corresponding to the top 3% is z = 1.881
Then (12.60 – x)/0.24 = 1.881  x = 12.1488
13.1208 ounces
13.0512 ounces
12.0792 ounces
12.1488 ounces
13) The tread life of a particular brand of tire is a random variable best described by a
normal distribution with a mean of 60,000 miles and a standard deviation of 1200 miles.
What warranty should the company use if they want 96% of the tires to outlast the
warranty?
The z-value that corresponds to the lower 4% (another way of saying that 96%
will last longer) is -1.751.
Then (x – 60,000)/1200 = -1.751  x = 57,898.80
57,900 miles
62,100 miles
61,200 miles
58,800 miles
14) Use a normal probability plot to asses whether the sample data could have come from
a population that is normally distributed.
An industrial psychologist conducted an experiment in which 40 employees that were
identified as "chronically tardy" by their managers were divided into two groups of size
20. Group 1 participated in the new "It's Great to be Awake!" program, while Group 2
had their pay pay docked. The following data represent the number of minutes that
employees in Group 1 were late for work after participating in the program.
not normally distributed
normally distributed
15) Use a normal probability plot to asses whether the sample data could have come from
a population that is normally distributed.
The following data represent a random sample of the number of shares of a
pharmaceutical company's stock traded for 20 days in 2000.
normally distributed
not normally distributed