Download Math 235 Answers (Practice-Sept

Math 235 Answers (Practice-Sept. 05)
Answers to Additional Problems—click here
1.
a)
40C5
= 40 * 39 * 38 * 37 * 36 Ans. 1/658,008
5!
= 658,008
b)
40P5
= 40*39*38*37*36 = 78,960,960 Ans. 1/78,960,960
2. Mean 69.8458
Stan. Dev. 4.1628 (pop.) 4.1824 (sd)
3. P(r > 2) = 1 – P(r < 2) = 1 – [ P(0) + P(1) ] = 1 – [ (½)5 + 5 * (½)5 ] = 13/16
4. P (heart and spade) = P (heart on 1st, spade on 2nd) + P (heart on 2nd, spade on 1st)
(Reason: the events are mutually exclusive)
a) With replacement:
¼*¼+¼*¼=⅛
b) Without replacement:
1 12 1 12 3 3
6
2
*  *
 


4 51 4 51 51 51 51 17
5. 6 * 10 * 4 = 240
6. The coefficient of variation (CV) for Group A is 2.3/35.4 * 100 = 6.5%. For Group B
the CV is 1.4/8.1 * 100 = 17.3%
So Group B has values which are more spread out.
7. –
8.
x
0
1
2
3
4
Total
P(x)
1/16
¼
3/8
¼
1/16
16/16 = 1
Answers to Additional Problems
1. a) Use the formula: Probability of r successes on n trials is: nCr (p)r (q)n–r
Note n = 5, p = .45, q = .55, and r varies from 0 to 5.
r
P(r)
0.0503
0
0.2059
1
0.3369
2
0.27565
3
0.1128
4
0.0185
5
b) The fastest way is to use the formulas
μ = np and σ = √(npq)
Then μ = 5 ·0.45 = 2.25,
σ = (5)(.45)(.55) = 1.1124
Alternately, you can work the values out using the formulas for mean and s.d.—see # 4
below.
c) For a probability dist., the mean is called the expected value. The 2.25 represents the
most likely number of heads obtained if the coin is tossed 5 times. I.e., if you repeat the
experiment (tossing the coin five times) a large number of times, the number of heads
obtained will average 2.25.
2. Use the same formula as in #1a:
P(r = 10) = 20C10 (.7)10(.3)10 = 184,756 (.7)10(.3)10 = .0308 (about 3%)
3. The prob. of getting any one problem correct is 1/5 or 0.2, because the student is
guessing. Thus p = 0.2, q = 0.8. There are 15 problems so n = 15.
Note P(r > 3) = 1 – P(r < 3) = 1 – [ P(0) + P(1) + P(2) ]
P(0) 15 C 0  (0.2) 0  (0.8)15  .0352
P(1) 15 C1  (0.2)1  (0.8)14  .1319
P(2) 15 C 2  (0.2) 2  (0.8)13  .2309
So the total probability is 1 – (.0352 + .1319 + .2309) =.602 (about 60%).
4.
Age,
years
Percent of
Patients
Midpoint
20-29
30-39
40-49
50-59
60-69
70-79
Totals
.02
.14
.18
.23
.24
.19
1
24.5
34.5
44.5
54.5
64.5
74.5
Midpoint ·
Frequency:
x·P(x)
.49
4.83
8.01
12.535
15.48
14.155
55.5
Deviation
Dev.2
Dev.2 ·
Freq.
-31
-21
-11
-1
9
19
961
441
121
1
81
361
19.22
61.74
21.78
0.23
19.44
68.59
191
The mean is μ = Σ [x · P(x)] = 55.5
The variance is σ2 = Σ[(x – μ)2 · P(x)] = 191 and the stan. dev. is √191 = 13.8203
5. We have r = number of calls out of n resulting in a sale, n = total number of calls, p =
.23 (prob. that any one call will result in a sale).
We want the smallest integer n such that
P(r > 0) > .99
i.e.,
1 – P(r = 0) > .99
1 – .99 > P(r = 0)
.01 > P(r = 0)
.01 > (.23)n
log (.01) > log (.23)n
-2 > n log (.23)
-2 > n ·-.6383
(-2 / -.6383) < n
(change the direction of the inequality since
dividing by a negative)
3.133 < n
The answer is 4 since that is the first integer satisfying 3.133 < n. Note that (.23)3 = .012
which is too large.
Four calls are needed to ensure with a probability of at least 99% that one or more of the
calls will result in a sale.