PROBABILITY & STATISTICAL
METHODS AND APPLICATIONS
5.23 The customer service department for a wholesale electronics outlet claims that 90 percent of all customer complaints are resolved to the satisfaction of the customer. In order to test this claim, a random sample of 15 customers who have filed complaints is selected.
a. Let x=the number of sampled customers whose complaints were resolved to the customer’s satisfaction. Assuming the claim is true, write the binomial formula for this situation.
X ̴B (x, 15, 0.9)
f(x) =
= 0, otherwise
b. b Use the binomial tables (see Table A.1, page 783) to find each of the following if we assume that the claim is true:
(1) P(x < 13). = 0.451
(2) P(x > 10). = 0.987
(3) P(x > 14). = 0.206
(4) P(9<x<12). = 0.184
(5) P(x < 9). = 0.002
c. Suppose that of the 15 customers selected, 9 have had their complaints resolved satisfactorily. Using part b, do you believe the claim of 90 percent satisfaction? Explain.
No, the claim of 90 percent satisfaction is not true because the P (x≤9) = 0.002 which is very small.
METHODS AND APPLICATIONS
5.30 Suppose that x has a Poisson distribution with u= 2.
a Write the Poisson formula and describe the possible values of x.
X ̴ P (x, 2)
f(x) =
x can take any integer values greater than or equal to 0
b Starting with the smallest possible value of x, calculate p(x) for each value of x until p(x) becomes smaller than .001.
|
x |
p(x) |
|
0 |
0.135 |
|
1 |
0.271 |
|
2 |
0.271 |
|
3 |
0.180 |
|
4 |
0.090 |
|
5 |
0.036 |
|
6 |
0.012 |
|
7 |
0.003 |
|
8 |
0.001 |
c Graph the Poisson distribution using your results of b.
d FindP(x=2). = 0.271
g Find P(x>1)and P(x>2). , P(x>1) = 0.865, P(x>2) = 0.323
i Find P(2<x<5). = 0.271
e FindP(x<4). = 0.947
h Find P(1<x<4) = 0.812.
j Find P(2<x<6) = 0.589.
f Find P(x<4) = 0.947.
METHODS AND APPLICATIONS
6.6 Suppose that the random variable x has a uniform distribution with c = 2 and d = 8.
a Write the formula for the probability curve of x, and write an interval that gives the possible values of x.
X ̴ U (2, 8)
f (x) = 1/6; 2 ≤x ≤ 8
= 0, otherwise
b Graph the probability curve of x.
f(x)
1/6
2 X 8
c FindP(3<x<5). = 1/3
d Find P(1.5 < x < 6.5). = 0.75
e Calculate the mean μ, variance σ2, and standard deviation σX.
mean = (c+d)/2 = 5
variance = ( c-d)^2)/12 = 3
standard deviation = 1.7321
f Calculate the interval [μ + 2σX]. What is the probability that x will be in this interval?
P (1.5358 ≤ X
6.26 Suppose that the random variable x is normally distributed with meanμ= 1,000 and standard deviation σ2=100. Sketch and find each of the following probabilities:
a P(1,000<x<1,200) = 0.4772
b P(x>1,257) = 0.0051
c P(x < 1,035) = 0.6368
d P(857<x<1,183) = 0.89
e P(x<700) = 0.0013
f P(812<x<913) = 0.1621
g P(x > 891) = 0.8621
h P(1,050<x<1,250) = 0.3023
6.54 Suppose that the random variable x has an exponential distribution with λ= 2.
a Write the formula for the exponential probability curve of x. What are the possible values of x?
f (x) =1/ 2
b Sketch the probability curve.
c Find P(x < 1). = 0.803
d Find P(.25<x<1).= 0.276
e Find P(x > 2). = 0.368
f Calculate the mean,μx , the variance, σx2, and the standard deviation, σx, of the exponential distribution of x.
mean = ½ = 0.5
variance = ¼ = 0.25
standard deviation = 0.5
g Find the probability that x will be in the interval [μx + 2σx].
P(-0.5 < x< 1.5) = 0.528