Download Probability problems, set 2

Shaver Manufacturing, Inc. offers dental insurance to its employees. A recent study by the
human resource director shows the annual cost per employee per year followed the normal
probability distribution, with a mean of $1,280 and a standard deviation of $420 per year.
What fraction of the employees cost more than $1,500 per year for dental expenses?
The interval for x > $1,500 must be standardized using the formula for z
x
220
 0.524

420
Since this z value represents the area of the normal curve from the extreme left to 0.524 to the
right of the normal curve midpoint. Hence, the answer to the question is:
Z = 1.00 – 0.524 = 0.476 = 0.48
Looking up that z value in the table for area under the normal curve, look for z = 0.4, move to
the right to the column marked “8”, which gives:
P = .3156 or 31.56% of the total Shaver Mfg. employees.
Since the question was posed as a fraction, the answer is “nearly one-third”.
z=

What fraction of the employees cost between $1,500 and 2,000 per year?
First, solve for z for each of the amounts.
We know from the above that $1,500 = z of 0.524
Solving for $2,000, we get:
z=
x


2000  1500 500

 1.1905
420
420
Next, solve for the probability interval:
P (1.191 ≥ z ≥ 0.524)
Find the corresponding areas under the normal curve:
P =(0.5+0.1170) – (0.5+0.3015) = 0.617 - .8015 = 0.1845 (ignore the sign or just transpose
the order in the original P formula)
P= 0.1845 = 18.45% = “Less than one-sixth”
Sorry, my first answer was off.
Estimate the percent that did not have any dental expense.
If you try solving for both z values, you get zero. So use $1 as the cut-off point
z=
x

1  1280
1

 0.0024
420
420

Proceeding:
1-(0.50 - .0024)*2 = 1-(0.4976)*2 = 1-0.9952 = 0.5%
Answer: one-half of 1 percent.
What was the cost for the 10 percent of employees who incurred the highest dental expenses?
x  $1280  1.26  ≥ $1,809.20
$420
7.42 The accounting department at Weston Materials, Inc., a national manufacturer of
unattached garages, reports that it takes two construction workers a mean of 32 hours and a
standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow
the normal distribution.
1. Determine the Z value for 29 and 34 hours. What percent of the garages take between
32 hours and 34 hours to erect?
z = 29 hours
x   29  32 3

  1.5
z=

2
2
z = 34 hours
x   34  32 2

  1.0
z=

2
2
Proceeding and looking up the corresponding values in the z table
P (-1.5 ≥ z ≥ +1.0) = (0.50-0.0668)+(0.5+0.3413) = 0.8413-.4332 = 0.4081 = 40.81%
2. What percent of the garages take between 29 hours and 34 hours to erect?
Question duplicates #1
3. What percent of the garages take 28.7 hours or less to erect?
z=
x


28.7  32 3.3

 1.65
2
2
P ≤ 1-(0.50 + 1.65) = 1-51.65 = 1 – 0.7422 = 0.2578 = 25.8%
4. Of the garages, 5 percent take how many hours or more to erect?
z=
x


X  32
 1.61 
2
THERE WAS A SLIGHT MISCALC IN THE PREVIOUS VERSION
X=35.2 hours
Because 1.61 is the z value where p becomes 0.9463. The operative phrase in the question is,
“…or more…” hence, you look at the top end of the normal curve, where the area cuts off at
95%.
6.62Suppose 1.5 percent of the antennas on new Nokia cell phones are defective. For a
random sample of 200 antennas, find the probability that:
a. None of the antennas is defective.
This is a binomial distribution:
(p is the probability of "successes"; in this case, a success is a defective cell
phone.)
(q is the probability of "failures"; in this case, a failure is a working
cell phone)
we're asked for a sample of 200...
..what is the probability that zero...
...cell phones are defective. We just plug everything into the binomial distribution
probability formula..
and we get..
b.Three or more of the antennas are defective.
Since 1.5% defect rate is the same as three or more antennas being defective, the answer is
the same.
3.82 The following frequency distribution reports the electricity cost for a sample of 50 twobedroom apartments in Albuquerque, New Mexico during the month of May last year.
Electricity Cost Frequency
$ 80 up to $1003
100 up to 120 8
120 up to 140 12
140 up to 160 16
160 up to 180 7
180 up to 200 4
Total
50
a.Estimate the mean cost.
Electricity
Cost
$ 80 up to
$100
100 up to 120
Freq- Low of
uency CI
3
Midpoint
fX
12
140 up to 160
160 up to 180
180 up to 200
Total
16
7
4
50
∑fX/n
s
s2
80
100
90.5
272
100
120
110.5
884
120
140
130.5
1,566
$51.20
$31.20
$11.20
140
160
150.5
2,408
$8.80
$77.44
160
180
170.5
1,194
$28.80
$829.44
180
200
190.5
762
$48.80
$2,381.44
7,085
-7
7,009
8
120 up to 140
Mean =
High of
CI
$2,621.44
$973.44
$125.44
$141.70
b.Estimate the standard deviation.
(x) 2
x 
n
S2 =
n 1
7009
7
50
=
50  1
 7  140.2
=
49
= 147.73/49 = -3.003
2
c.Use the Empirical Rule to estimate the proportion of costs within two standard deviations of
the mean. What are these limits?
P (1SD < x < 1SD) = (.5-.1587)+(.5+.1587) = -0.3413 + 0.6587 =
Lower limit = 141.70 – 3.003 = $138.697
Upper limit = 141.70 + 3.003 = $144.703