Question

Math 3023                                                                  Homework # 04                                                                Spring 2018

                                                                           Prairie View A & M University

Name: ___________________________

(1). The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between 0 and 15 minutes, inclusive.

(a). On the average, how long must a person wait? [Hint: Find the mean (expected value)]

(b). Find the standard deviation of the r.v.?

(c). Ninety percent of the time, the time a person must wait falls below what value?

(2). Suppose that the amount of time one spends in a bank is exponentially distributed with the parameter .

=

(a). What is the probability that a customer will spend more than 15 minutes in the bank?

(b). Find the mean of the random variable.

(c). Compute the standard deviation of the random variable.

(3). Suppose the weight of a specific fish species is normally distributed with mean of 10 pounds and standard deviation of 2 pounds. If a fish of this species is caught randomly, what is the probability of the following events?

(a). Weight of the fish is less than 10 pounds?

(b). Weight of the fish is in between 7 pounds and 8.5 pounds?

(c). If two fish are caught randomly, what is the probability that the weights of both fish will be more than 9 pounds?

(d). Calculate the 80^th percentile of weights of this type of fish.

Answer 1

1)
a)
mean = (a+b)/2 = (0+15)/2 = 7.5
On the average a person miust wait 7.5 minutes

b)
std.dviation = sqrt(b-a)^2/12 =sqrt((15-0)^2/12) = 4.3

c)
let m = 0.90

P(X<m) = base *height
0.90 = (m-0)*1/15
m = 13.5
Ninety percent of the time, the time a person must wait falls below 13.5 value

2)
a)
here, lambda = 1/10
P(X>15) = e^-15lambda
= e^(-15*1/10) = 0.22

b)
mean = 1/lambda = 1/(1/10) = 10

c)
variance = 1/(lambda)^2
= 100
std.deviation = sqrt(variance) = 10

3)
mean = 10 , s = 2

a)
P(x<10)
z =( x -mean)/s
= ( 10 - 10)/2
= 0
P(x <10) = P(z <0) = 0.5 by using standard normal table

b)
P(7 < x < 8.5)
= P((7 -10)/2 < z < (8.5 -10)/2)
= P(-1.5 < z < -0.75)
P(7 < x < 8.5) = P(-1.5 < z < -0.75) = 0.1598by using standard normal table

c)
p(x >9) when n =2
z =( x -mean)/s
= ( 9 - 10)/(2/sqrt(2))
= -0.7071

P(x>9) = P(z <-0.7071) = 0.3107
by using standard normal table

d)
z value at 80 % = 0.8416

z =( x -mean)/s
0.8416 = ( x -10)/2
x = 11.6832