# Chapter-6-Section-1-Day-2-Continuous-Random-Variables-2016s-Notes

```Chapter 6 Section 1 Day 2 2016s.notebook
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Honors Statistics
Aug 23-8:26 PM
3. Review OTL C6#2
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6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
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A Skip 4, 12, 16
Write a program to generate random
numbers. I've decided to give them
free will.
Apr 25-10:55 AM
Toss 4 times Suppose you toss a fair coin 4 times.
Let X = the number of heads you get.
First List the Sample Space ....
HHHH
THHH
HHHT
THHT
HHTH
THTH
HHTT
THTT
HTHH
TTHH
HTHT
TTHT
HTTH
TTTH
HTTT
TTTT
(a) Find the probability distribution of X.
(b) Make a histogram of the probability distribution. Describe what you see.
frequency
0.5
0.4
0.3
0.2
0.1
1
0
2
3
4
(c) Find P(X ≤ 3) and interpret the result.
P( X ≤ 3) =
+
+
+
=
15
16
= 0.9375
The probability that 4 tosses of a coin results in 3 or fewer heads is 0.9375
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Kids and toys In an experiment on the behavior of young
children, each subject is placed in an area with five toys. Past
experiments have shown that the probability distribution of
the number X of toys played with by a randomly selected
subject is as follows:
> (a) Write the event “plays with at most two toys” in terms of X.
> What is the probability of this event?
P(x ≤ 2) = 0.03 + 0.16 + 0.30 = 0.49
> (b) Describe the event X > 3 in words.
>
The probability that the young child plays with more than 3 toys.
>
What is its probability?
>
What is the probability that X ≥ 3? P(X ≥ 3) = 0.28 + 0.23 = 0.51
P(X > 3) = 0.17 + 0.11 = 0.28
Nov 28-12:08 AM
Kids and toys Refer to Exercise 4.
Calculate the mean of the random
variable X and interpret this result in
context.
µx = 0(0.03) + 1(0.16) + 2(0.30) + 3(0.23) + 4(0.17) + 5(0.11) = 2.68
If many, many children participated in this experiment, the
mean number of toys that randomly selected children would
play with will average 2.68 toys.
(The expected number of toys a randomly selected young child
will play with is 2.68.) This statement is optional.
Nov 28-12:16 AM
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Kids and toys Refer to Exercise 4. Calculate and interpret the standard deviation
of the random variable X. Show your work.
σ2x = (0-2.68)2(0.03) + (1-2.68)2(0.16)
+ (2-2.68)2(0.30) + (3-2.68)2(0.23)
+ (4-2.68)2(0.17) + (5-2.68)2(0.11) = 1.7176
σx =√1.7176 = 1.31057
The standard deviation of X is σx = 1.31057
The number of toys a randomly selected
young child will play with will typically differ
from the mean (2.68) by about 1.31 toys.
Nov 28-12:22 AM
Benford’s law Faked numbers in tax returns, invoices, or
expense account claims often display patterns that aren’t
present in legitimate records. Some patterns, like too many
round numbers, are obvious and easily avoided by a clever
crook. Others are more subtle. It is a striking fact that the
first digits of numbers in legitimate records often follow a
model known as Benford’s law.7 Call the first digit of a
randomly chosen record X for short. Benford’s law gives this
probability model for X (note that a first digit can’t be 0):
(a) Show that this is a legitimate probability distribution.
all individual probabilities are between 0 and 1
0.301 + 0.176 + 0.125 + 0.097 + 0.079 + 0.067 + 0.058 + 0.051 + 0.046 = 1
(b) Make a histogram of the probability distribution. Describe what you see.
See histogram above. The distribution is NOT symmetric. It is skewed
to the right. The data should be analyzed using the 5 number summary.
(c) Describe the event X ≥ 6 in words. What is P(X ≥ 6)?
What is the probability that the first digit in a legitimate legal document is 6 or greater?
P(X ≥ 6) = 0.067 + 0.058 + 0.051 + 0.046 = 0.222
(d) Express the event “first digit is at most 5” in terms of X. What is the probability
of this event?
P(X < 6) = 1 - P(X ≥ 6) = 1 - 0.222 = 0.778
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Benford’s law Refer to Exercise 5. The first digit of a randomly chosen expense
account claim follows Benford’s law. Consider the events A = first digit is 7 or
greater and B = first digit is odd.
(a) What outcomes make up the event A? What is P(A)?
P(X ≥ 7) = 0.058 + 0.051 + 0.046 = 0.155
(b) What outcomes make up the event B? What is P(B)?
P(X is odd) = 0.301 + 0.125 + 0.079 + 0.058 + 0.046 = 0.609
(c) What outcomes make up the event “A or B”? What is P(A or B)? Why is this
probability not equal to P(A) + P(B)?
P(X ≥ 7 or X is odd) = 0.609 + 0.155 - (0.058 + 0.046) = 0.66
Both 7 and 9 are included in each event and must their sum must be subtracted
because they were counted twice ( the general probability addition rule)
Nov 28-12:14 AM
Benford’s law and fraud A not-so-clever employee decided to fake his monthly
expense report. He believed that the first digits of his expense amounts should be
equally likely to be any of the numbers from 1 to 9. In that case, the first digit Y of
a randomly selected expense amount would have the probability distribution
shown in the histogram.
> (a) Explain why the mean of the random variable Y is located at the solid red line
in the figure.
The mean is the balance point of the distribution. So it is located in the
center of a uniform or symmetric distribution histogram in this case at 5.
> (b) The first digits of randomly selected expense amounts actually follow
Benford’s law (Exercise 5). According to Benford’s law, what’s the expected value
of the first digit? Explain how this information could be used to detect a fake
expense report.
µx = 1(0.301) + 2(0.176) + 3(0.125) + 4(0.097) + 5(0.079)
+ 6(0.067) + 7(0.058) + 8(0.051) + 9(0.046) = 3.441
To detect a fake expense report, compute the sample mean of the first
digits of the numbers on the report. A value closer to 3.441 suggests a
truthful report but a value closer to 5 (the more uniform distribution)
suggest a false report.
> (c) What’s P(Y > 6) in the above distribution? According to Benford’s law, what
proportion of first digits in the employee’s expense amounts should be greater
than 6? How could this information be used to detect a fake expense report?
P(Y > 6) = 0.058 + 0.051 + 0.046 = 0.155
For a uniform distribution the P(Y > 6) = 0.3
To detect a fake expense report, compute the percent of the first digits
that are greater than 6 on the report. A value closer to 0.155 suggests a
truthful report but a value closer to 0.3 (the more uniform distribution)
suggest a false report.
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Benford’s law and fraud Refer to Exercise 13. It might also be possible to detect an
employee’s fake expense records by looking at the variability in the first digits of
those expense amounts.
> (a) Calculate the standard deviation σY. This gives us an idea of how much
variation we’d expect in the employee’s expense records if he assumed that
first digits from 1 to 9 were equally likely.
σ2x = (1-5)2(0.10) + (2-5)2(0.10) + (3-5)2(0.10)
+ (4-5)2(0.10) + (5-5)2(0.10) + (6-5)2(0.10)
+ (7-5)2(0.10) + (8-5)2(0.10) + (9-5)2(0.10) = 6.66
σx =√6.66 = 2.58
> (b) Now calculate the standard deviation of first digits that follow Benford’s
law (Exercise 5). Would using standard deviations be a good way to detect
fraud? Explain.
σ2x = (1-3.44)2(0.301) + (2-3.44)2(0.176) + (3-3.44)2(0.125)
+ (4-3.44)2(0.097) + (5-3.44)2(0.079) + (6-3.44)2(0.067)
+ (7-3.44)2(0.058) + (8-3.44)2(0.051) + (9-3.44)2(0.046)
= 6.06052
σx =√6.06052 = 2.42
Because the standard deviations are so close 2.58 and
2.42 it would be difficult to determine fake reports from
legitimate reports using the standard deviation.
Nov 28-12:22 AM
Pair-a-dice Suppose you roll a pair of fair, six-sided dice.
Let T= the sum of the spots showing on the up-faces.
(a) Find the probability distribution of T.
Probability model for the sum of two fair dice
(b) Make a histogram of the probability distribution.
Describe what you see.
0.16
0.12
0.08
0.04
2
3
4
5
6
7
8
9
10
11
12
(c) Find P(T ≥ 5) and interpret the result.
(d) Find the expected value of the distribution. Interpret this value in context.
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Pair-a-dice Suppose you roll a pair of fair, six-sided dice.
Let T= the sum of the spots showing on the up-faces.
(a) Find the probability distribution of T.
Probability model for the sum of two fair dice
frequency of sum
(b) Make a histogram of the probability distribution.
Describe what you see.
0.16
0.12
0.08
0.04
2
3
4
5
6
7
8
9
10
11
12
sum of dice
(c) Find P(T ≥ 5) and interpret the result.
P(T ≥ 5) = 1 - P(T < 5) = 1 - (
1
+
36
2
36
+
3
36
)=1-
6
36
=
30
36
= 0.8333
The probability that a random roll of dice will add to more
than 5 is 0.8333 or 83.3%
(d) Find the expected value of the distribution. Interpret this value in context.
1
2
3
4
36
36
36
5
µT = 2(36 ) + 3( ) + 4( ) + 5( ) + 6( 36) + 7(
5
4
3
2
1
6
36
)+
252
8( 36) + 9(36 ) + 10( ) + 11( 36) + 12(36 ) = 36 = 7
36
If a pair of fair dice are rolled many, many times and the sum
of the dots is calculated the sum will average a total of 7.
The expected sum of a pair of randomly rolled dice is 7.
Nov 27-10:56 PM
You choose a 3 digit number. The lottery commission announces the
The "box" pays \$83.33 if the number you choose has the same digits as
on the box.
(Assume a number with three different digits is chosen)
let X =
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Area above the x axis
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The probability distribution for a continuous random variable assigns probabilities to intervals of
outcomes rather than to individual outcomes. In fact, all continuous probability models assign
probability 0 to every individual outcome. Only intervals of values have positive probability. To see that
this is true, consider a specific outcome from the random number generator of the previous example,
such as P(Y = 0.7). The probability of this event is the area under the density curve that’s above the
point 0.70000…on the horizontal axis.But this vertical line segment has no width, so the area is 0. For
that reason,
UT I O
N
CA
P(0.3 ≤ Y ≤ 0.7) = P(0.3 ≤ Y < 0.7) = P(0.3 < Y < 0.7) = 0.4
ALL continuous probability models assign
probability 0 to every individual outcome.
P(x=3) = 0
In many cases, discrete random variables arise from counting something
—for instance, the number of siblings that a randomly selected student has.
Continuous random variables often arise from measuring something
—for instance, the height or time to run a mile for a randomly selected student.
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b) P(X
HOMEWORK PROBLEM
d) What important fact about continuous random variables does comparing
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CLASS EXAMPLE
Weights of 3-year-old females
The weights of 3-year-old females closely follow a Normal Distribution with a mean of
µ = 30.7 pounds and a standard deviation of σ = 3.6 pounds. Randomly choose
one 3-year-old female and call her weight X.
What is the probability that a randomly selected 3-year-old female weights at least 30
pounds?
What is the probability that a randomly selected 3-year-old female weights between 28
and 33 pounds?
Dec 2-3:52 PM
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Discuss with students to guide them when doing #20 in the homework assignment
Apr 27-9:24 AM
Owned Units:
The histogram of owned housing units is symmetric. It is
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May 7-8:36 AM
7.5 x = the number of rooms in a randomly chosen
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Working out Choose a person aged 19 to 25 years at random and ask, “In the past seven
days, how many times did you go to an exercise or fitness center or work out?” Call the
responseY for short. Based on a large sample survey, here is a probability model for the
(a) Show that this is a legitimate probability distribution.
0.68 + 0.05 + 0.07 + 0.08 + 0.05 + 0.04 + 0.01 + 0.02 = 1
frequency
(b) Make a histogram of the probability distribution. Describe what you see.
0.5
0.4
0.3
0.2
0.1
0
1
2
3
5
6
7
Number of workout days
(c) Describe the event Y < 7 in words. What is P(Y < 7)?
What is the probability that a randomly selected person aged
19 to 25 went to the gym less than seven days this week?
P(Y < 7) = 1 - P(Y = 7) = 1 - 0.02 = 0.98
(d) Express the event “worked out at least once” in terms of Y. What is the probability of this event?
P(Y ≥ 1) = 1 - P(Y = 0) = 1 - 0.68 = 0.32
Nov 29-10:57 AM
27, 28, 29, 30
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Working out Refer to Exercise 6. Consider the events A = works out at least once
and B = works out less than 5 times per week.
(a) What outcomes make up the event A? What is P(A)?
outcomes = 1,2,3,4,5,6,7
P(Y ≥ 1) = 1 - P(Y = 0) = 1 - 0.68 = 0.32
(b) What outcomes make up the event B? What is P(B)?
outcomes = 0,1,2,3,4
P(Y < 5) = 0.68 + 0.05 + 0.07 + 0.08 + 0.05 = 0.93
(c) What outcomes make up the event “A and B”? What isP(A and B)? Why is this
probability not equal to P(A) ·P(B)?
P(A and B) = 0.05 + 0.07 + 0.08 + 0.05 = 0.25
The events working out at least once and working out less than 5 times per week
are not INDEPENDENT events. So Multiplication cannot be used to determine
the probability of P(A and B)
Nov 29-11:00 AM
Keno Keno is a favorite game in casinos, and similar games are popular with the
states that operate lotteries. Balls numbered 1 to 80 are tumbled in a machine as the
bets are placed, then 20 of the balls are chosen at random. Players select numbers by
marking a card. The simplest of the many wagers available is “Mark 1 Number.” Your
payoff is \$3 on a \$1 bet if the number you select is one of those chosen. Because 20 of
80 numbers are chosen, your probability of winning is 20/80, or 0.25. Let X= the net
amount you gain on a single play of the game.
OR
Based on what you "get" back
OR
µx = -1(0.75) + 2(0.25) = -0.25
Nov 29-11:02 AM
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Size of American households In government data, a household consists of all
occupants of a dwelling unit, while a family consists of two or more persons who live
together and are related by blood or marriage. So all families form households, but
some households are not families. Here are the distributions of household size and
family size in the United States:
Legitimate Distributions ...
Let X = the number of people in a randomly selected U.S. household and Y = the
number of people in a randomly chosen U.S. family.
(a) Make histograms suitable for comparing the probability distributions of X and
Y. Describe any differences that you observe.
Dec 5-5:48 PM
`
> (b) Find the mean for each random variable. Explain why this difference makes sense.
µx = 1(0.25) + 2(0.32) + 3(0.17) + 4(0.15) + 5(0.07) + 6(0.03) + 7(0.01) = 2.6
If many, many random households were selected, they would
on average have 2.6 people dwelling in them.
µy = 1(0.0) + 2(0.42) + 3(0.23) + 4(0.21) + 5(0.09) + 6(0.03) + 7(0.02) = 3.14
If many, many random families were selected, they would on
average have 3.14 people in the family.
> (c) Find and interpret the standard deviations of both Xand Y.
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> (c) Find and interpret the standard deviations of both X and Y.
σ2x = (1-2.6)2(0.25) + (2-2.6)2(0.32) + (3-2.6)2(0.17)
+ (4-2.6)2(0.15) + (5-2.6)2(0.07) + (6-2.6)2(0.03)
(7-2.6)2(0.01) = 2.02
σx =√2.02 = 1.421
The standard deviation of X is σx = 1.421
The number of people in a randomly
selected household will typically differ from
the mean (2.6) by about 1.421 people.
σ2y = (1-3.14)2(0.0) + (2-3.14)2(0.42) + (3-3.14)2(0.23)
+ (4-3.14)2(0.21) + (5-3.14)2(0.09) + (6-3.14)2(0.03)
(7-3.14)2(0.02) = 1.5604
σx =√1.5604 = 1.249
The standard deviation of X is σy = 1.249
A the number of people in a randomly
selected family will typically differ from the
mean (3.14) by about 1.249 people.
Dec 4-7:00 AM
THIS IS ON THE CLASS WORKSHEET !
Random numbers Let Y be a number between 0 and 1 produced by a random
number generator. Assuming that the random variable Y has a uniform
distribution, find the following probabilities:
HOMEWORK PROBLEM
#22 Random numbers
a) P(Y≤ 0.4) = (0.4)(1) = 0.40
b) P(Y < 0.4) = (0.4)(1) = 0.40
c) P(0.1 < Y ≤ 0.15 or 0.77 ≤ Y < 0.88) = (0.05)(1) + (0.11)(1) = 0.16
d) What important fact about continuous random variables does comparing
The area of a line segment (Y = 0.4) is equal to zero
Nov 29-11:16 AM
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Running a mile A study of 12,000 able-bodied male students at
the University of Illinois found that their times for the mile run
were approximately Normal with mean 7.11 minutes and
standard deviation 0.74 minute.10 Choose a student at random
from this group and call his time for the mile Y. Find P(Y < 6)
and interpret the result.
N(7.11, 0.74)
This interprets (in the context of this problem)... The probability of
randomly choosing a random student who can run the mile in less
than 6 minutes is approximately 6.68% or 6.7 out of 100.
Nov 29-11:19 AM
Ace! Professional tennis player Rafael Nadal hits the ball extremely hard. His
first-serve speeds follow a Normal distribution with mean 115 miles per hour
(mph) and standard deviation 6 mph. Choose one of Nadal’s first serves at
random. Let Y = its speed, measured in miles per hour.
N(115,6)
(a) Find P(Y > 120) and interpret the result.
≈ 0.2033
This interprets (in the context of this problem)... The probability of
randomly choosing one of Nadal's first serves that is faster than 120
mph is approximately 20.33% or 20 out of 100.
(b) What is P(Y ≥ 120)? Explain. The answer is equal to P(Y > 120) ≈ 0.2033
Because the P(Y = 120) is zero.
Nov 29-11:21 AM
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(c) Find the value of c such that P(Y ≤ c) = 0.15. Show your work.
0.15
c=?
c = -1.04(6) + 115
z = -1. 04
c = 108.76
This interprets (in the context of this problem)... The probability of
randomly choosing one of Nadal's first serves that is slower than
108.76 mph is approximately 15% or 15 out of 100.
Dec 1-9:51 PM
Pregnancy length The length of human pregnancies from conception to birth
follows a Normal distribution with mean 266 days and standard deviation 16
days. Choose a pregnant woman at random. Let X = the length of her
pregnancy.
N(266, 16)
x = 240
(a) Find P(X ≥ 240) and interpret the result.
≈ 1 - 0.0516 ≈ 0.9484
This interprets (in the context of this problem)... The probability of
randomly choosing one pregnant lady that carries her baby at least
240 days is approximately 94.84% or 95 out of 100 pregnant women.
(b) What is P(X > 240)? Explain.
The answer is equal to P(X > 240) ≈ 0.9484
Because the P(X = 240) is zero.
Dec 4-7:12 AM
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(c) Find the value of c such that P(X ≥ c) = 0.20. Show your work.
N(266, 16)
0.20
Look up 0.80
so this
side is 0.80
c=?
z = 0. ??
c = 0.84(16) + 266
c = 279.44
This interprets (in the context of this problem)... The probability of
randomly choosing one pregnant lady that carries her baby at least
279.44 days is approximately 20% or 20 out of 100 pregnant women.
Dec 4-7:12 AM
Multiple choice: Select the best answer for Exercises 27 to 30.
Exercises 27 to 29 refer to the following setting. Choose an American household at
random and let the random variable X be the number of cars (including SUVs and light
trucks) they own. Here is the probability model if we ignore the few households that own
more than 5 cars:
What’s the expected number of cars in a randomly selected American household?
(a) 1.00
(b) 1.75
(c) 1.84
(d) 2.00
(e) 2.50
µx = 0(0.09) + 1(0.36) + 2(0.35) + 3(0.13) + 4(0.05) + 5(0.02) = 1.75
Dec 4-7:12 AM
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The standard deviation of X is σX = 1.08. If many households were selected at
random, which of the following would be the best interpretation of the value
1.08?
(a) The mean number of cars would be about 1.08.
(b) The number of cars would typically be about 1.08 from the mean.
(c) The number of cars would be at most 1.08 from the mean.
(d) The number of cars would be within 1.08 from the mean about 68% of the
time.
(e) The mean number of cars would be about 1.08 from the expected value.
Dec 4-7:12 AM
C
About what percentage of households have a number of cars within 2
standard deviations of the mean?
(a) 68%
we know the following ....
µx = 1.75
(b) 71%
(c) 93%
(d) 95%
(e) 98%
σx = 1.08
but it is not stated that the distribution
is approximately NORMAL so use the table
above ....
mean + 2 St. Dev. = 1.75 + 2(1.08) = 3.91
mean - 2 St. Dev. = 1.75 - 2(1.08) = -0.41
between 0 and 3.91 cars (must use 3)
0.09 + 0.36 + 0.35 + 0.13 = 0.93
What is the probability that they have
more than 5 cars ?????
Dec 4-11:45 AM
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ITBS scores The Normal distribution with mean µ = 6.8 and standard
deviation σ = 1.6 is a good description of the Iowa Test of Basic Skills
(ITBS) vocabulary scores of seventh-grade students in Gary, Indiana.
Call the score of a randomly chosen student X for short. Find P(X ≥ 9)
and interpret the result.
N(6.8, 1.6)
≈ 0.0838
This interprets (in the context of this problem)... The probability of
randomly choosing a random seventh grade student who scores a 9
or higher on the IOWA test is approximately 8.38% or 8.4 out of 100.
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Dec 4-11:22 AM
Honors Statistics
POP Quiz
Chapter 6 Section 1
Nov 1-1:41 PM
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Spell-checking Spell-checking software catches
“nonword errors,” which result in a string of letters
that is not a word, as when “the” is typed as “teh.”
essay (without spell-checking), the numberX of
nonword errors has the following distribution:
(a) Write the event “at least one nonword error” in terms of X. What is the
probability of this event?
1) = .2 + .3 + .3 + .1 = .9
(b) Describe the event X ≤ 2 in words. What is its probability? What is the
probability that X < 2?
what is the probability that two or less nonword errors are caught?
P(x ≤ 2) = .1 + .2 + .3 = .6
P(x < 2 ) = .1 +.2 = .3
c. Find the expected value of the distribution. Interpret this value in context.
µx = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.3) + 4(0.1) = 2.1
selected, the number of non word error will average 2.1 per
The expected number of nonword errors in a randomly
selected undergraduate 250 word essay is 2.1.
d. Find the standard deviation of the distribution. Interpret this value in context.
σ2x = (0-2.1)2(0.1) + (1-2.1)2(0.2)
+ (2-2.1)2(0.3) + (3-2.1)2(0.3)
+ (4-2.1)2(0.1) = 1.29
σx =√1.29 = 1.136
The standard deviation of X is σx = 1.136
A randomly selected undergrad essay will
typically differ from the mean (2.1) by about
1.136 nonword errors.
Nov 21-12:23 PM
N(42,2)
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Sep 20-8:20 PM
ITBS scores The Normal distribution with mean µ = 6.8 and standard
deviation σ = 1.6 is a good description of the Iowa Test of Basic Skills
(ITBS) vocabulary scores of seventh-grade students in Gary, Indiana.
Call the score of a randomly chosen student X for short. Find P(X ≥ 9)
and interpret the result.
N(6.8, 1.6)
This interprets (in the context of this problem)... The probability of
randomly choosing a random seventh grade student ....
Nov 29-11:20 AM
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A deck of cards contains 52 cards, of which 4 are aces. You are offered the following
wager: Draw one card at random from the deck. You win \$10 if the card drawn is an
ace. Otherwise, you lose \$1. If you make this wager very many times, what will be the
mean amount you win?
µx = -1(
48
52
) + 10(
4
52
) = -0.1538
(a) About − \$1, because you will lose most of the time.
(b) About \$9, because you win \$10 but lose only \$1.
(c) About −\$0.15; that is, on average you lose about 15 cents.
(d) About \$0.77; that is, on average you win about 77 cents.
(e) About \$0, because the random draw gives you a fair bet.
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