In: Statistics and Probability
1) A discrete random variable X has the following probability distribution:
x | 13 | 18 | 20 | 24 | 27 |
p(x) | 0.22 | 0.25 | 0.2 | 0.17 | 0.16 |
Compute each of the following quantities (give the exact numbers):
(a) P(18)=
(b) P(X>18)=
(c) P(X?18)=
(d) The mean ? of X is
(e) The variance ?2 of X is (write all decimal points)
(f) The standard deviation ? of X is (use 4 decimal points)
2) Borachio works in an automotive tire factory. The number X of sound but blemished tires that he produces on a random day has the probability distribution
x | 2 | 3 | 4 | 5 |
p(x) | 0.48 | 0.36 | 0.12 | 0.04 |
For the following, don’t round the numbers
(a) Find the probability that Borachio will produce more than three blemished tires tomorrow.
(b) Find the probability that Borachio will produce at most two blemished tires tomorrow.
(c) Compute the mean of X.
3)Find the mean ? and standard deviation ? of a random variable X with the following probability distribution
x | 0 | 1 | 2 | 3 |
p(x) | 0.3 | 0.3 | 0.3 | 0.1 |
(a) ?=1.5, ?=0.96
(b) ?=1.2, ?=0.96
(c) ?=1.5, ?=0.98
(d) ?=1.2, ?=0.98
(e) ?=1.5, ?=2.4
4). To set a schedule for a new bus route, the transit authority repeatedly times the trip between two points; the time X in minutes is found to have the following probability distribution function
x | 20 | 21 | 22 | 23 | 24 | 25 |
p(x) | 0.07 | 0.19 | 0.36 | 0.24 | 0.1 | 0.04 |
(a) the probability that a randomly selected trip will take at least 24 minutes is about:
(i) 0.14; (ii) 0.10; (iii) 0.23; (iv) 0.04; (v) 0.86
(b) The probability that a randomly selected trip will take between 21 and 24 minutes (including 21 and 24) is about: (i) 0.19; (ii) 0.79; (iii) 0.89; (iv) 0.29; (v) 0.7
(c) If the trip is made over and over, the average time it takes in minutes is about:
(i) 22.5; (ii) 22.2; (iii) 21.9; (iv) 22.0; (v) 23.0
5) A sociologist surveyed the households in a small town. The random variable X represents the number of dependent children in the households. The missing probability is
x | 0 | 1 | 2 | 3 | 4 |
p(x) | 0.07 | 0.2 | 0.38 | ? | 0.13 |
(a) 0.02
(b) 0.12
(c) 0.22
(d) 0.32
(e) 0.78
6) The following is a probability distribution function. True or False?
x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
p(x) | 0.3 | 0.25 | 0.25 | 0.1 | 0.05 | 0.03 | 0.02 |
x | 0 | 1 | 2 | 3 | 4 | 5 |
p(x) | 3/4 | 1/10 | 1/20 | 1/25 | 1/50 | -1/100 |
x | 0 | 1 | 2 | 3 |
p(x) | 0.005 | 0.435 | 0.555 | 0.206 |
x | 0 | 1 | 2 | 3 | 4 |
p(x) | 0.05 | 0.25 | 0.35 | 1.25 | -0.9 |
x | 0 | 1 | 2 | 3 | 4 |
p(x) | 0.17 | 0.1 | 0.35 | 0.2 | 0.16 |
1)
a)
P(18) = 0.25
b)
P(x >18) = P(20) + P(24) + P(27)
= 0.2 + 0.17 + 0.16 = 0.53
c) P(x <=18) = P(13) + P(18)
= 0.22 + 0.25 = 0.47
d)
mean = 13 *0.22 + 18 * 0.25 = 20 * 0.2 + 24 * 0.17 + 27 *
0.16
= 19.760
e)
Variance = ( 13 - 19.760)^2 * 0.22 + (18- 19.760)^2 * 0.25 + (20-
19.760)^2 * 0.2 + (24- 19.760)^2 * 0.17 + (27- 19.760)^2 * 0.16
Variance = 22.28240
f)
std.deviation = sqrt(variance)
= sqrt(22.28240)
= 4.7204
2)
a)
P(x > 3) =P(4) + P(5)
= 0.12 + 0.04 = 0.16
b)
P(x < =2) = P(2) = 0.48
c)
mean = 2 * 0.48 + 3 * 00.36 + 4 * 0.12 + 5 * 0.04
= 2.72000
3)
mean = 0 * 0.3 + 1 * 0.3 + 2 * 0.3 + 3 * 0.1
= 1.2
std.deviation = (0-1.2)^2 * 0.3 + (1-1.2)^2 * 0.3 + (2-1.2)^2 *
0.3 + (3-1.2)^2 * 0.1
= 0.96
std.deviation = sqrt(0.96) = 0.98
Option d)
4)
a)
P(x >=24) = P(x =24) + P(x =25)
= 0.1 + 0.04 = 0.14
b)
P(21 < x < 24)
= P(21) + P(22) + P(23) + P(24)
= 0.19 + 0.36 + 0.24 + 0.1
= 0.89
c)
average = 20 * 0.07 + 21* 0.19 + 22 * 0.36 + 23 * 0.24 + 24 * 0.1 +
25 * 0.04
= 22.23