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MATH 217 Test 2 Version A

Name: KEY Sec Number:
Answer all questions to the best of your ability. Note you should show as much work as is possible.
For questions answered using Excel be sure to include the Excel ’code’ on your exam to ensure full
credit. Complete credit will not be earned for just answers.
1. (2 pts) A man has six shirts and four ties. Assuming that they all match how many different
shirt-and-tie combinations can he wear?
The man has (6)(4) = 24 shirt and tie options.
2. (2 pts) A woman has five blouses and three skirts. Assuming that they all match, how many
different outfits can she wear?
The woman has (5)(3) = 15 blouse and skirt combinations.
3. (2 pts) Suppose you are going to burn a compact disc (CD) that will contain 12 songs. In
how many ways can you arrange the 12 songs on the CD?
There are 12! or 479001600 possible ways to arrange the songs.
4. (8 pts) Given the following diagram with circles A, B, and C. The probability of each region
is shown.
0.05
A B
C
0.35
0.4
0.15 0.05
Find the following probabilities:
(a) P(A|C):
(b) P(Ac
|B):
(c) P((A and B)|B):
(d) P(not A|C
c
):
P(A|C) = 0
P(A
c
|B) = 0.45/0.6 = 0.75
P((A and B)|B) = 0.15/0.6 = 0.25
P(not A|C
c
) = 0.4
0.9
= 4/9 = 0.444444444
5. (12 pts) In the 2004 baseball season, Ichiro Suzuki of the Seattle Mariners set the record for
the most hits in a season with a total of 262 hits. In the following probability distribution,
the random variable X represents the number of hits Ichiro obtained in a game.
x P(x) x P(x)
0 0.1677 3 0.1491
1 0.3354 4 0.0373
2 0.2857 5 0.0248
(a) Verify that this is a discrete probability distribution.
0.1677 + 0.3354 + 0.2857 + 0.1491 + 0.0373 + 0.0248 = 1
(b) Compute the mean and standard deviation.
This is a weighted average:
µX = 0.1677(0)+0.3354(1)+0.2857(2)+0.1491(3)+0.0373(4)+0.0248(5) = 1.6273
σX =
X
(x − µX)
2P(x)

0.5
= 1.178471345
(c) Interpret the mean of the random variable.
Ichiro Suzuki is expected to get 1.6237 hits per game.
(d) What is the probability that in a randomly selected game Ichiro got 2 hits?
The probability that Ichiro gets 2 hits in a randomly selected game is 0.2857.
(e) What is the probability that in a randomly selected game Ichro got more than 1 hit?
The probability that Irico gets more than one hit is 0.4969.
P(2) + P(3) + P(4) + P(5) = 0.496
6. (8 pts) The number of hits to a Web site follows a Poisson process; hits occur at the rate of
1.4 per minute between 7:00pm and 9:00pm. Compute the probability that the number of
hits between 7:30pm and 7:35pm is:
(a) Exactly seven
P(7) = ((1.4)(5))7
7! e
−1.5(5) = 0.14900278
The probability of exactly seven hits to the web site is 0.149.
(b) Fewer than three
P(< 3) = P(0)+P(1)+P(2) = 0.000911882+0.006383174+0.022341108 = 0.029636164
The probability that there is less than three hits is 0.029636.
(c) Greater than or equal to three Here we use the complement of the previous question
yielding 0.970363836.
P(≥ 3) = 1 − P(< 3) = 1 − 0.029636 = 0.970363836
7. (8 pts) In March 1995, the Harris Poll reported that 80% of parents spank their children.
Suppose a recent poll o 1030 adult Americans with children finds that 781 indicated that
they spank their children. If we assume parents’ attitude toward spanking has not changed
since 1995, how many of the 1030 parents surveyed would we expect to spank? Do the results
suggest parents’ attitude toward spanking may have changed since 1995? Why?
Here we have a binomial distribution with mean and standard deviation given by:
µX = np = (1030)(0.8) = 824 and σX =
p
np(1 − p) = (1030(0.8)(0.2))1/2 = 12.837445
We would expect 824 of the parents surveyed to spank their children. The value 781 has a
z−score of
z − score =
781 − 1030
12.837445
= −3.349576
this is more than two standard deviations below the mean suggesting that the observation
would be an outlier and perhaps people have changed there attitude toward spanking.
8. (10 pts) The number of Chocolate chips in an 18-ounce package of cookies is approximately
normally distributed with a mean of 1262 chips and standard deviation of 118 chips.
(a) What is the probability that a randomly selected 18-ounce package of cookies contains
between 1000 and 1400 chocolate chips (inclusively).
Z =
1000 − 1262
118
= −2.22033898 and Z =
1400 − 1262
118
= 1.169491525
P(X ≤ 1000) = P(Z ≤ −2.22) = 0.0132 and P(X ≤ 1400) = P(Z ≤ 1.17) = 0.8790
P(1000 ≤ X ≤ 1400) = 0.879 − 0.0132 = 0.8658
The probability of having between 1000 and 1400 chips in a bag of cookies is 86.58
percent.
(b) What is the probability that a randomly selected package of cookies contains fewer than
1000 chips? Using some of the work from above the probability of having a bag of cookies
contain fewer than 1000 chips is 0.0132 or 1.3 %.
(c) What portion of cookie packages contain more than 1200 chips?
Z =
1200 − 1262
118
= −0.5254237
P(X ≥ 1200) = P(Z ≥ −0.5254) = 1 − P(Z ≤ −0.5254) = 1.0 − 0.2981 = 0.7019
The probability that a bag has more than 1200 chips is 0.7019.