Download 1) Given independent random variables with means and standard

1) Given independent random variables with means and standard deviations as shown, find the mean and
standard deviation of:
a)2Y + 20
b) 3X
c) 0.25X + Y
d) X – 5Y
e) X1 + X2 + X3
X
Y
Mean
80
12
SD
12
3
SOLUTION to #1:
a) 𝐸(2𝑌 + 20) = 2𝐸(𝑌) + 20 = 44
b) 𝐸(3𝑋) = 3𝐸(𝑋) = 240
𝑆𝐷(2𝑌 + 20) = 𝑆𝐷(2𝑌) = 2𝑆𝐷(𝑌) = 6
𝑆𝐷(3𝑋) = 3𝑆𝐷(𝑋) = 36
c) 𝐸(0.25𝑋 + 𝑌) = 0.25𝐸(𝑋) + 𝐸(𝑌) = 32
𝑆𝐷(0.25𝑋 + 𝑌) = √0.252 122 + 32 = √18 = 4.2426
d) 𝐸(𝑋 − 5𝑌) = 𝐸(𝑋) − 5𝐸(𝑌) = 20
𝑆𝐷(𝑋 − 5𝑌) = √122 + 52 32 = √369 = 19.2094
e) 𝐸(𝑋1 + 𝑋2 + 𝑋3 ) = 𝐸(𝑋1 ) + 𝐸(𝑋2 ) + 𝐸(𝑋3 ) = 3𝐸(𝑋) = 240
𝑆𝐷(𝑋1 + 𝑋2 + 𝑋3 ) = √122 + 122 + 122 = 20.7846
2) Justin Thyme needs to get from from Anytown to Cityville. The quickest way for Justin to do so is to take a
train from Anytown to Boondocksburg, and then from Boondocksburg to Cityville. The travel time for the train
from Anytown to Boondocksburg varies normally with mean 127 minutes and standard deviation 8 minutes,
while the travel time for the train from Boondocksburg to Cityville varies normally with mean 93 minutes and
standard deviation 4 minutes. The layover in Boondocksburg (i.e., the time Justin has to wait for the next train)
is 45 minutes. Justin leaves the train station in Anytown at noon and is supposed to meet his friend at the train
station in Cityville by 4:15. What is the probability he’ll make it in time?
SOLUTION to #2: The ultimate question is this … what is the probability that the total travel time will be less
than 4 hours and 15 minutes (i.e., 255 minutes)? In symbols, we want 𝑃(𝑇 ≤ 255). In order to answer this
question we need to define some variables.
Let X = travel time from A to B (which varies normally with mean and standard deviation as given above)
Let Y = travel time from B to C (which also varies normally with mean and standard deviation as given above)
Let T = X + Y + 45 (which is the sum of the two travel times and the layover in Boondocksburg – note that T will
also vary because X and Y both vary.)
Remember that our goal is to find 𝑃(𝑇 ≤ 255). To answer probability questions about T, we need to know its
distribution. Since T is the sum of a constant and two random variables that each follow normal distributions,
we know that T will also be normally distributed with a mean of 𝑬(𝑻) = 𝑬(𝑿) + 𝑬(𝒀) + 𝟒𝟓 = 𝟐𝟓𝟓 and a
standard deviation of 𝑺𝑫(𝑻) = √𝟖𝟐 + 𝟒𝟐 = √𝟖𝟎 = 𝟖. 𝟗𝟒𝟒𝟑.
Now that we know the distribution of T, we can answer probability questions by using z-scores and the standard
normal table or the normalcdf command on our calculators.
255 − 265
𝑃(𝑇 ≤ 255) = 𝑃 (𝑍 ≤
) = 𝑃(𝑍 ≤ −1.1180) = 0.1318
8.9443
So, there’s about a 13% chance that Justin will make it to Cityville in time to meet his friend.
3) The probability model below describes the number of repair calls that an appliance repair shop may receive
during an hour.
a)
b)
c)
d)
e)
Repair Calls
0
1
2
3
4
5
Probability
0.1 0.11 0.24 0.35 0.13 0.07
How many calls should the shop expect per hour?
What is the variance and the standard deviation?
What is the probability that the shop will have at most 3 repair calls during a randomly selected hour?
What is the probability that the number of repair calls will be within 1 SD of the expected number?
If the number of repair calls is independent from hour to hour, what is the probability that the shop will
have a total of at most 6 calls during two randomly selected hours?
SOLUTION to #3:
a) Multiply each value by its associated probability and add the results. (0)(0.1) + (1)(0.11) + … = 2.51.
b) Multiply the squared deviation of each value from the mean by its associated probability and add the
results. The variance is VAR(X) = (0-2.51)2(0.1) + (1-2.51)2(0.11) + (2-2.51)2(0.24) + … = 1.7499 and the
standard deviation is the square root of the variance, or SD(X) = 1.3228.
c) 𝑃(𝑋 ≤ 3) = 0.1 + 0.11 + 0.24 + 0.35 = 0.80. Or, noting that only 4 and 5 are NOT included, we could
more easily calculate 𝑃(𝑋 ≤ 3) = 1 − (0.13 + 0.07) = 0.80.
d) Note that 𝜇 − 𝜎 = 2.51 − 1.32 = 1.19 and 𝜇 + 𝜎 = 2.51 + 1.32 = 3.83. Both 2 and 3 are between these
values, so the desired probability is P(2) + P(3) = 0.24 + 0.35 = 0.59.
e) We need to determine the probabilities of obtaining anywhere from 0 to 6 calls over two independently
selected hours. Here are the possibilities and their probabilities.
No. of calls
0
1
2
3
4
5
6
Possibilities
Probability
00 (this means 0 1st hour, 0 2nd hour) (.1)(.1)=0.01
01 or 10
(.1)(.11)(2) = 0.022
02 or 11 or 20
(.1)(.24)(2) + (.11)(.11)=0.0601
03, 12, 21, 30
(.1)(.35)(2)+(.11)(.24)(2)=0.1228
04, 13, 22, 31, 40
(.1)(.13)(2)+(.11)(.35)(2)+(.24)(.24)=0.1606
05, 14, 23, 32, 41, 50
[(.1)(.07)+(.11)(.13)+(.24)(.35)](2)=0.2106
15, 24, 33, 42, 51
[(.11)(.0 7)+(.24)(.13)](2)+(.35)(.35)=0.2003
𝑷(𝒙 ≤ 𝟔) = 𝟎. 𝟕𝟖𝟔𝟒 (the sum of the above probabilities)
Hmmm … perhaps it would have been faster to take 1 – P(7 or more)?
4) The communications monitoring company Postini has reported that 91% of e-mail messages are “spam” (i.e.,
junk mail). Recently, you installed a spam filter. You observe that over the past week the filter only allowed 151
of 1422 e-mails to pass through the filter, classifying the rest as junk. Should you worry that the filtering is too
aggressive?
a) Explain why we can assume this situation follows a binomial setting.
b) Use a normal approximation to the binomial to calculate the probability implied by the question. Be sure to
verify that the conditions that allow the use of the normal approximation are met.
SOLUTION to #4:
a) We have two outcomes (“junk” or not), the probability of email being called “junk” is the same from trial to
trial, the trials are considered independent of one another, and we’re counting the number of “junk” emails we
get in a fixed number of trials (n=1422).
b) Define X as “the number of emails classified as junk.” Note that 𝑛𝑝 = .91(1422) = 1294.02 and 𝑛𝑞 =
𝑛(1 − 𝑝) = 1422(. 09) = 127.98 are both larger than 10, so using a normal approximation is reasonable. The
normal approximation will have mean and standard deviation 𝜇 = 𝑛𝑝 = 1294.02 and 𝜎 = √𝑛𝑝𝑞 = 10.7917. If
only 151 emails got through, then the filter is classifying 1422 – 151 = 1271 as junk. Is this unreasonable if 91%
of emails really are junk? In other words, what is 𝑃(𝑋 ≥ 1271)? Find the z-score: 𝑃 (𝑍 ≥
1271−1294
)
10.7912
=
𝑃(𝑍 ≥ −2.133) = 0.9835. This means that it is 98% likely that at least 1271 of our 1422 emails really ARE junk,
so I’d conclude the filter is not being overly aggressive.
5) Assume that 13% of people are left-handed. If we select 5 people at random, find the probability of each of
the following.
a) There are some lefties among the people chosen
b) There are exactly 3 lefties in the group
c) There are at least 3 lefties in the group
d) There are no more than 3 lefties in the group
SOLUTION to #5:
Let X = # of lefties in the sample of 5.
a) This is the same as asking for the probability that there are more than 0 lefties, i.e., P(X=1, 2, 3, 4, or 5). This
is easiest to calculate by determining the probability that X is not 0. In other words, we want 1 – P(X=0), or in
other words (symbols?), 1 – P(all righties) = 1 – 0.875 = 0.5016.
b) 𝑃(𝑋 = 3) = (53)(. 13)3 (. 87)2 = 0.0166
c) 𝑃(𝑋 ≥ 3) = 𝑃(𝑋 = 3) + 𝑃(𝑋 = 4) + 𝑃(𝑋 = 5) = (53)(. 13)3 (. 87)2 + (54)(. 13)4 (. 87)1 + (. 13)5 = 0.0179
(or use the binomcdf command … we need to get rid of P(X=0, 1, or 2), so 1-binomcdf(5,.13,2)).
d) 𝑃(𝑋 ≤ 3) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3)
5
5
5
= (. 87)5 + ( ) (. 13)1 (. 87)4 + ( ) (. 13)2 (. 87)3 + ( ) (. 13)3 (. 87)2 = 0.9987
1
2
3
(or use the binomcdf command … we need P(X=0, 1, 2, or 3) = binomcdf(5,.13,3))