Download Test2c

STP 226
NAME:
Fall 2000
ID:
Exam #2C Key
This is a closed book exam. You may use a calculator and a formula page from the
book.
1. Given P(A) = 0.7 and P(notB) = 0.6 and P(AorB) = 0.8
a) (5 points) compute P(B)
P(B)=1-.6=.4
b) (6 points) Compute P (A&B).
.8 = .7 + .4 -P(A&B), P(A&B) = .3
2. Toss a fair coin twice.
a) (5 points) Let A=first toss came up heads.
List outcomes in A.
A={ htt, hht, hth, hhh}
b) (6 points) What is the probability that first toss is not a head?
P(notA) = 1 - 4/8 =.5
3. A frequency distribution for the number of siblings of 45 students in one of the
STP 226 classes is given in the table below:
Number of 0
1
2
3
4
5
6
7
siblings
Number of 7
6
12
8
5
4
2
1
students
(frequency)
For a student selected at random from that class, let
A= event that student has at most 2 siblings
B= event that student has at least 5 siblings
a. (7 points) Compute P(AorB)
32/45 = .711
b. ( 6 points) Are events A , B mutually exclusive? Clearly explain why or why not?
Yes, there are no common outcomes.
4.
Use the standard normal curve to find the following:
a. (7 points)
area between - 1.82 and 1.78
Area left of 1.78 - area left of -1.82 = .9625 - .0344 = .9281
b. (7 points)
z-value with the area equal to 0.004 to the left of it.
Z= -2.65
5.
The lifetime of a particular brand of fuse is normally distributed
with mean  = 1600 hours and standard deviation  = 50 hours. Let X be
the lifetime of randomly selected fuse of that brand.
a. (7 points) Sketch the distribution of that population and point out all
of its main features (shape, center, spread)
Bell shaped curve, symmetric,centered at 1600, most area between 1450 and
1750, total area=1
b. (7 points) Determine the following probability:
P(X < 1680)= P(z < (1680-1600)/50) = P(z < 1.6) = .9452
b. (8 points) Derermine the 90th percentile of the distribution of X .
90th percentile of N(0, 1) is z=1.28. (area to the right of z=1.28 is 10 %)
x=+z, so x=1600+1.28(50) =1664
6. A random sample of size 25 was taken from the normally distributed population
with mean  = 150 and standard deviation  =20 .
Let x denote the mean of this sample.
a. (6 points) Name the sampling distribution of x , give the mean and the
standard deviation of x .
Normal distribution  x =150,  x =4
N(150,4)
b.
(7 points)Compute the probability that x is less than 153:
P ( x < 153) =P(z< (53-150)/4) = P(z<.75) = .7734
7. You want to estimate the true mean weight of all the students at the large
university (  )
A random sample of n = 9 students gives a sample mean of 145 lb.
Assume that the weights of the students at that university are normally
distributed with standard deviation of 18 lb.
a. (6 points) Obtain 90% confidence interval for , the true mean
weight.
Z-interval, since  is known
145+/- 1.645(18/3) gives (135.13, 154.87)
b. (5points)Looking at your interval, is the true mean likely to be
 = 153 lb.? Clearly explain your answer.
Yes, since clearly 153 is in the interval and we have 90% confidence that this
interval captures the true mean.
c. (5points)What sample size do you need so that your confidence interval has a
margin of error of no more than 4 lb. (keep the same 90% confidence level).
n=[(1.6458*18)/4]2 =54.79 ~ 55