Schaum's Outline Probability and Statistics (3rd Edition)

142 CHAPTER 4 Special Probability Distributions

Miscellaneous problems
4.54. The probability that an entering college student will graduate is 0.4. Determine the probability that out of

5 students (a) none, (b) 1, (c) at least 1, will graduate.

(a) P(none will graduate) ϭ 5C0(0.4)0(0.6)5 ϭ 0.07776, or about 0.08
(b) P(l will graduate) ϭ 5C1(0.4)1(0.6)4 ϭ 0.2592, or about 0.26
(c) P(at least 1 will graduate) ϭ 1 Ϫ P(none will graduate) ϭ 0.92224, or about 0.92

4.55. What is the probability of getting a total of 9 (a) twice, (b) at least twice in 6 tosses of a pair of dice?

Each of the 6 ways in which the first die can fall can be associated with each of the 6 ways in which the second
die can fall, so there are 6 ? 6 ϭ 36 ways in which both dice can fall. These are: 1 on the first die and 1 on the
second die, 1 on the first die and 2 on the second die, etc., denoted by (1, 1), (1, 2), etc.

Of these 36 ways, all equally likely if the dice are fair, a total of 9 occurs in 4 cases: (3, 6), (4, 5), (5, 4),
(6, 3). Then the probability of a total of 9 in a single toss of a pair of dice is p ϭ 4>36 ϭ 1>9, and the probability
of not getting a total of 9 in a single toss of a pair of dice is q ϭ 1 Ϫ p ϭ 8>9.

1 2 8 6Ϫ2 61,440
9 9 531,441
(a) P(two 9s in 6 tosses) ϭ 6C2 a b a b ϭ

(b) P(at least two 9s) ϭ P(two 9s) ϩ P(three 9s) ϩ P(four 9s) ϩ P(five 9s) ϩ P(six 9s)

1 2 8 4 1 3 8 3 1 4 8 2 1 5 8 1 2
9 9 9 9 9 9 9 9 9
ϭ 6C2 a b a b ϩ 6C3 a b a b ϩ 6C4 a b a b ϩ 6C5 a b ϩ 6C6 a b

ϭ 61,440 ϩ 10,240 ϩ 960 ϩ 48 ϩ 1 ϭ 72,689
531,441 531,441 531,441 531,441 531,441 531,441

Another method

P(at least two 9s) ϭ 1 Ϫ P(zero 9s) Ϫ P(one 9)

1 0 8 6 1 1 8 5 72,689
9 9 9 9 531,441
ϭ 1 Ϫ 6C0 a b a b Ϫ 6C1 a b a b ϭ

4.56. If the probability of a defective bolt is 0.1, find (a) the mean, (b) the standard deviation for the distribu-
tion of defective bolts in a total of 400.

(a) Mean ϭ np ϭ 400(0.1) ϭ 40, i.e., we can expect 40 bolts to be defective.

(b) Variance ϭ npq ϭ 400(0.l)(0.9) ϭ 36. Hence the standard deviation ϭ !36 ϭ 6.

4.57. Find the coefficients of (a) skewness, (b) kurtosis of the distribution in Problem 4.56.

(a) Coefficient of skewness ϭ qϪp ϭ 0.9 Ϫ 0.1 ϭ 0.133
!npq 6

Since this is positive, the distribution is skewed to the right.

1 Ϫ 6pq 1 Ϫ 6(0.1)(0.9)
(b) Coefficient of kurtosis ϭ 3 ϩ npq ϭ 3 ϩ 36 ϭ 3.01

The distribution is slightly more peaked than the normal distribution.

4.58. The grades on a short quiz in biology were 0, 1, 2, . . . , 10 points, depending on the number answered cor-
rectly out of 10 questions. The mean grade was 6.7, and the standard deviation was 1.2. Assuming the
grades to be normally distributed, determine (a) the percentage of students scoring 6 points, (b) the max-
imum grade of the lowest 10% of the class, (c) the minimum grade of the highest 10% of the class.

(a) To apply the normal distribution to discrete data, it is necessary to treat the data as if they were continuous.
Thus a score of 6 points is considered as 5.5 to 6.5 points. See Fig. 4-20.

5.5 in standard units ϭ (5.5 Ϫ 6.7)>1.2 ϭ Ϫ1.0

6.5 in standard units ϭ (6.5 Ϫ 6.7)>1.2 ϭ Ϫ0.17

CHAPTER 4 Special Probability Distributions 143

Required proportion ϭ area between z ϭ Ϫ1 and z ϭ Ϫ0.17
ϭ (area between z ϭ Ϫ1 and z ϭ 0)
Ϫ (area between z ϭ Ϫ0.17 and z ϭ 0)
ϭ 0.3413 Ϫ 0.0675 ϭ 0.2738 ϭ 27%

Fig. 4-20 Fig. 4-21

(b) Let x1 be the required maximum grade and z1 its equivalent in standard units. From Fig. 4-21 the area to the
left of z1 is 10% ϭ 0.10; hence,

Area between z1 and 0 ϭ 0.40
and z1 ϭ Ϫ1.28 (very closely).

Then z1 ϭ (x1 Ϫ 6.7)>1.2 ϭ Ϫ1.28 and x1 ϭ 5.2 or 5 to the nearest integer.

(c) Let x2 be the required minimum grade and z2 the same grade in standard units. From (b), by symmetry,
z2 ϭ 1.28. Then (x2 Ϫ 6.7)>1.2 ϭ 1.28, and x2 ϭ 8.2 or 8 to the nearest integer.

4.59. A Geiger counter is used to count the arrivals of radioactive particles. Find the probability that in time t
no particles will be counted.

Let Fig. 4-22 represent the time axis with O as the origin. The probability that a particle is counted in a small
time ⌬t is proportional to ⌬t and so can be written as ␭⌬t. Therefore, the probability of no count in time ⌬t is
1 Ϫ ␭⌬t. More precisely, there will be additional terms involving (⌬t)2 and higher orders, but these are
negligible if ⌬t is small.

Fig. 4-22

Let P0(t) be the probability of no count in time t. Then P0(t ϩ ⌬t) is the probability of no count in time
t ϩ ⌬t. If the arrivals of the particles are assumed to be independent events, the probability of no count in
time t ϩ ⌬t is the product of the probability of no count in time t and the probability of no count in time ⌬t.
Therefore, neglecting terms involving (⌬t)2 and higher, we have

(1) P0(t ϩ ⌬t) ϭ P0(t)[l Ϫ ␭⌬t]

From (1) we obtain lim P0(t ϩ ⌬t) Ϫ P0(t) ϭ ϪlP0(t)
(2) ⌬t
⌬t S0

i.e.,

(3) dP0 ϭ ϪlP0 or dP0 ϭ Ϫl dt
dt P0

Solving (3) by integration we obtain

ln P0 ϭ Ϫ␭t ϩ c1 or P0(t) ϭ ceϪ␭t

To determine c, note that if t ϭ 0, P0(0) ϭ c is the probability of no counts in time zero, which is of course 1.
Thus c ϭ 1 and the required probability is

(4) P0(t) ϭ eϪ␭t