Download 1. The mean life of a computer disk drive is 2,000 hours. The

1.
The mean life of a computer disk drive is 2,000 hours. The standard deviation is 140 hours. Assuming the
life-time of the drives to be normally distributed, find the probability of a disk drive lasting more than 1,800
hours.
a. .4236
b. .9236
c.
.8472
d.
.5764
e.
.2118
f.
None of the above.
z = (x – )/ = (1800 – 2000)/140 = -1.42857
P(x > 1800) = P(z > -1.42857) = 0.9236.
2.
The yearly commissions per salesperson employed by a manufacturer of light machinery average $40,000
with a standard deviation of $5,000. What percent of the salespersons earn between $32,000 and $42,000?
a.
60.06%
b.
39.94%
c.
34.13%
d.
65.87%
e.
81.66%
f.
None of the above.
z = (x – )/ = (32000 – 40000)/5000 = -1.6 and z = (42000 – 40000)/5000 = 0.4
P(32000 < x < 42000) = P(-1.6 < z < 0.4) = 0.6006
This is 60.06%.
3.
A study of 200 stamping firms revealed these incomes after taxes:
Income After Taxes Number of Firms
Under $1 million 102
$1 million up to $20 million
$20 million and more
61
37
What is the probability that a particular firm selected has under $1 million in income after taxes?
a.
0.000
b.
0.250
c.
0.333
d.
0.510
e.
None of the above.
P(Under 1 million) = 102/200 = 0.510.
4.
A study of the opinion of designers with respect to the primary color most desirable for use in executive
offices showed that:
Primary Color Number of Opinions
Red 92
Orange 86
Yellow 46
Green 91
Blue 37
Indigo 46
Violet
2
What is probability that a designer does not prefer red?
a.
1.000
b.
0.770
c.
0.000
d.
0.230
e.
None of the above.
Total number of designers = 400
Number of designers who do not prefer red = 400 – 92 = 308
P(A designer does not prefer red) = 308/400 = 0.770
5.) The manufacturer of salad dressings uses machines to dispense the dressing into bottles that move along an
assembly line. The machine is working well when 8 ounces is dispensed. The standard deviation of the process
is 0.15 ounce. A sample of 50 bottles is selected periodically and the assembly line is stopped when there is
evidence that the average amount dispensed is less than 8 ounces. Suppose that a sample of 50 bottles reveals
an average of 7.983 ounces.
1. State the null and alternative hypotheses.
2. At the .05 level of significance, is there evidence that the average amount dispensed is less than 8 ounces?
H0: The average amount dispensed is 8 ounces
Ha: The average amount dispensed is less than 8 ounces
Lower-tailed z- test for population mean with  = 0.05
Lower Critical value of z = -1.645
Decision Rule: Reject H0 if z- score for the sample < -1.645
SE = s/n = 0.15/50 = 0.0212
z = (x-bar – )/SE = (7.983 – 8)/0.0212 = -0.8014
Since -0.8014 > -1.645, we fail to reject H0
There is no evidence that the average amount dispensed is less than 8 ounces.
6.) A machine being used for packaging raisins has been set so that, on average, 15 ounces of raisins will be
packaged per box. The operations manager wishes to test the machine setting and selects a sample of 30
consecutive raisin packages filled during the production process. The sample mean is 15.18 and the sample
standard deviation is 0.4909. Use the t-distribution. Is there evidence that the mean weight per box is different
from 15 ounces? (use level of significance equal to .05)
H0: The mean weight per box is 15 ounces
Ha: The mean weight per box is not 15 ounces
Two-tailed t- test for population mean with  = 0.05
Degrees of freedom = n – 1 = 30 – 1 = 29; Critical value of t = 2.0452
Decision Rule: Reject H0 if |t- score| for the sample > 2.0452
SE = s/n = 0.4909/30 = 0.08963
t = (x-bar – )/SE = (15.18 – 15)/0.08963 = 2.008
Since 2.008 < 2.0452, we fail to reject H0
There is no evidence that the mean weight per box is different from 15 ounces.