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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Honors Statistics
Aug 23-8:26 PM
3. Review OTL C6#6
emphasis Normal Distributions
Aug 23-8:31 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Nov 21-8:16 PM
Rainy days Imagine that we randomly select a day from the past 10 years. Let X be
the recorded rainfall on this date at the airport in Orlando, Florida, and Y be the
recorded rainfall on this date at Disney World just outside Orlando. Suppose that you
know the means µX and µY and the variances and of both variables.
(a) Is it reasonable to take the mean of the total rainfall X+ Y to be µX + µY? Explain
your answer.
It is reasonable to take the mean of the
total rainfall because the rainfall two
cities that are close in location does not
depend on independence. You can
always combine means of data sets.
(b) Is it reasonable to take the variance of the total rainfall to be Explain your answer.
It is NOT reasonable to take the variance of
the total rainfall because the rainfall of two
closely located cities will NOT be independent.
Dec 6-11:13 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Get on the boat! Refer to Exercise 41. Find the expected value and standard deviation
of the total amount of profit made on two randomly selected days. Show your work.
Big assumption ... two randomly selected days are
INDEPENDENT so .... go ahead and calculate!
Remember ...
µY = $19.35 - 20 = $-0.65
σY = $6.45
Use the formula T = Y1 + Y2
µT = $-0.65 + $-0.65 = $-1.30
Remember you can never add STANDARD DEVIATIONS.
σY = $6.45 so σ2Y = (6.45)2 = 41.6025
σ2T = σ2Y1 + σ2Y2
σ2T = 41.60 + 41.60 = 83.20
σT = √ 83.20 = $9.12
Dec 6-11:13 PM
Essay errors Typographical and spelling errors can be either “nonword errors” or “word errors.”
A nonword error is not a real word, as when “the” is typed as “teh.” A word error is a real word,
but not the right word, as when “lose” is typed as “loose.” When students are asked to write a
250-word essay (without spell-checking),
The number of nonword errors X in a randomly selected essay has the following probability
distribution:
The number of word errors Y has this probability distribution:
Assume that X and Y are independent.
An English professor deducts 3 points from a student’s essay score for each
nonword error and 2 points for each word error. Find the mean and standard
deviation of the total score deductions for a randomly selected essay. Show your
work.
Use Formula D = 3(X) + 2(Y)
D = deductions, X = "non-word error" and Y = word error
µD = 3(2.1) + 2(1.0) = 8.3 points deducted
The expected number of points deducted on two randomly
selected essays is 8.3 points.
If many, many essays are randomly selected, this is the
average amount of point deductions on two essays. (In the
long run !!)
Remember you can never add STANDARD DEVIATIONS.
σX = 1.136 so σ2X = (1.136)2 = 1.290496
σY = 1.0 so σ2Y = (1.0)2 = 1.0
σ2D = (3)2σ2X + (2)2σ2Y = 9(1.136)2 + 4(1.0)2 = 15.614464
σD = √ 15.614464 = 3.95451414 points
The standard deviation of the mean is 3.95 points.
The deductions on the 2 essays will typically differ
from the mean of 8.3 by 3.95 points.
Dec 6-11:14 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
(for word vs. non-word errors) deducted on a
... Fewer word errors than non-word errors
The results in the yellow triangle make up the event a randomly
selected student makes more word errors than non-word errors.
Dec 6-11:14 PM
women at a college has mean 120 and standard deviation 28, and the distribution of scores
The standard deviation of the mean is 44.82 points.
The amount of points in which the female outscores the males
From the information given, can you find the probability that the woman chosen scores
If we new the shape of the distribution (specifically if it
was normally distributed) then we could answer the
question. Currently we do not have enough information.
Dec 6-11:14 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Exercises 57 and 58 refer to the following setting. In Exercises 14 and
18 of Section 6.1,we examined the probability distribution of the random
variable X = the amount a life insurance company earns on a randomly
chosen 5-year term life policy. Calculations reveal that µX = $303.35 and
σX = $9707.57.
Life insurance The risk of insuring one person’s life is reduced if
we insure many people. Suppose that we insure two 21-year-old
males, and that their ages at death are independent. If X1 and X2
are the insurer’s income from the two insurance policies, the
insurer’s average income W on the two policies is
Find the mean and standard deviation of W.
(You see that the mean income is the same as for a single policy but the standard deviation is less.)
µW = $303.35 + $303.35 = $303.35
2
Remember you can never add STANDARD DEVIATIONS.
σX = $9707.57 so σ2x = (9707.57)2 = 94236915.3
σ2W = σ2X1 + σ2X2
2
σ2W = (9707.57)2 + (9707.57)2
22
2
σ2W = 47118457.65
σW = √ 47118457.65 = $6864.2885
Nov 30-7:46 PM
Life insurance If four 21-year-old men are insured, the insurer’s average income is
where Xi is the income from insuring one man. Assuming that the amount of income earned on
individual policies is independent, find the mean and standard deviation of V. (If you compare
with the results of Exercise 57, you should see that averaging over more insured individuals
reduces risk.)
µV = $303.35 + $303.35 + 303.35 + 303.35 = $303.35
4
Remember you can never add STANDARD DEVIATIONS.
σX = $9707.57 so σ2x = (9707.57)2 = 94236915.3
σ2V = σ2X1 + σ2X2 + σ2X3 + σ2X4
42
σ2V = (9707.57)2 + (9707.57)2 + (9707.57)2 + (9707.57)2
42
σ2V = 23559228.83
σV = √ 23559228.83 = $4853.785
Dec 6-11:13 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Nov 27-10:28 PM
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
Nov 27-9:53 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
new σ = old
σ
new σ = old
σ
σ
May 05, 2016
σ*b
σ
σ
σ
σ
σ2a+bx = b2*σ2
Dec 5-7:04 PM
=σ
=σ
Dec 5-7:26 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
=σ
=σ
Dec 5-7:25 PM
Nov 30-7:23 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Nov 30-7:23 PM
Dec 1-2:08 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
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Sep 26-6:57 PM
Sep 26-6:58 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
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Dec 9-9:54 AM
Dec 11-7:07 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
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Dec 11-7:05 PM
May 4-9:19 AM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Nov 21-8:16 PM
Time and motion A time-and-motion study measures the time required for an assembly-line worker to
perform a repetitive task. The data show that the time required to bring a part from a bin to its position on an
automobile chassis varies from car to car according to a Normal distribution with mean 11 seconds and
standard deviation 2 seconds. The time required to attach the part to the chassis follows a Normal
distribution with mean 20 seconds and standard deviation 4 seconds. The study finds that the times required
for the two steps are independent. A part that takes a long time to position, for example, does not take more
or less time to attach than other parts.
(a) What is the distribution of the time required for the entire operation of positioning and attaching a
randomly selected part?
T= P+A
µT = 11 + 20 = 31 minutes
Remember you can never add STANDARD DEVIATIONS.
σP = 2 so σ2P = (2)2 = 4
and σA = 4 so σ2A = (4)2 = 16
σ2T = 4 + 16 = 20
σ2T = σ2P + σ2A
σT = √ 20 = 4.472 minutes
(b) Management’s goal is for the entire process to take less than 30 seconds. Find the probability that this
goal will be met for a randomly selected part.
X:
17.59
22.06
26.53
31
35.47
39.94
44.41
y = 30
z=
4.47
= -0.2237
P( Y < 30 ) = P(z < -0.22) ≈ 0.4129
The probability that the management will meet the goal of less than 30 seconds for a randomly
selected part is approximately 41.29%
Dec 6-11:13 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Electronic circuit The design of an electronic circuit for a toaster calls for a 100-ohm
resistor and a 250-ohm resistor connected in series so that their resistances add. The
components used are not perfectly uniform, so that the actual resistances vary
independently according to Normal distributions. The resistance of 100-ohm resistors has
mean 100 ohms and standard deviation 2.5 ohms, while that of 250-ohm resistors has
mean 250 ohms and standard deviation 2.8 ohms.
X and Y are independent.
(a) What is the distribution of the total resistance of the two components in series for a
randomly selected toaster?
µ
σ
σ
(b) Find the probability that the total resistance for a randomly selected toaster lies
between 345 and 355 ohms.
The probability that the total resistance for a randomly selected toaster
lies between 345 and 355 ohms is approximately 81.64%
Dec 6-11:14 PM
Swim team Hanover High School has the best women’s swimming team in the region.
The 400-meter freestyle relay team is undefeated this year. In the 400-meter freestyle
relay, each swimmer swims 100 meters. The times, in seconds, for the four swimmers this
season are approximately Normally distributed with means and standard deviations as
shown. Assume that the swimmer’s individual times are independent. Find the probability
that the total team time in the 400-meter freestyle relay for a randomly selected race is
less than 220 seconds.
Use this formula
T=W+J+C+L
µT = 55.2 + 58 + 56.3 + 54.7 = 224.2
seconds
Remember you can never add STANDARD DEVIATIONS.
σJ = 3.0 so σ2J = (3.0)2 = 9.00
σW = 2.8 so σ2w = (2.8)2 = 7.84
σC = 2.6 so σ2C = (2.6)2 = 6.76
σL = 2.7 so σ2L = (2.7)2 = 7.29
σ2T = σ2W + σ2J + σ2C + σ2L
σ2T = 7.84 + 9.0 + 6.76 + 7.29 = 30.89
σT = √ 30.89 = 5.5578 or 5.56 seconds
X:
207.52
213.08
218.64
224.2
229.76
235.32
240.88
T = 220
z=
= -0.7553
5.56
P( T < 220 ) = P(z < -0.76) ≈ 0.2236
The probability that the total team time in the 400-meter freestyle relay
for a randomly selected race is less than 220 seconds is approximately 22.66% chance.
Dec 6-11:14 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Toothpaste Ken is traveling for his business. He has a new 0.85-ounce tube of toothpaste that’s
supposed to last him the whole trip. The amount of toothpaste Ken squeezes out of the tube each
time he brushes varies according to a Normal distribution with mean 0.13 ounces and standard
deviation 0.02 ounces. If Ken brushes his teeth six times on a randomly selected trip, what’s the
probability that he’ll use all the toothpaste in the tube?
T = D1 + D 2 + D3 + D4 + D5 + D6
Use this formula
µT = 0.13 + 0.13 + 0.13 + 0.13 + 0.13 + 0.13 = 0.78 ounces
Remember you can never add STANDARD DEVIATIONS.
σD = 0.02 so σ2D = (0.02)2 = 0.0004
σ2T = σ2D + σ2D + σ2D + σ2D + σ2D + σ2D
σ2T = 0.0004 + 0.0004 + .... + 0.0004
σT = √ 0.0024 = 0.048987 or 0.049 ounces
X:
0.633
0.682
0.731
0.78
0.829
0.878
0.927
T = 0.85
z=
= 1.4286
0.049
P( T > 0.85 ) = P(z < 1.43) ≈ 0.0764
1-.9236
The probability that Ken will use all the toothpaste in the tube is
approximately 7.64%
Dec 6-11:14 PM
Auto emissions The amount of nitrogen oxides (NOX) present in the exhaust of a particular type
of car varies from car to car according to a Normal distribution with mean 1.4 grams per mile (g/
mi) and standard deviation 0.3 g/mi. Two randomly selected cars of this type are tested. One has
1.1 g/mi of NOX; the other has 1.9 g/mi. The test station attendant finds this difference in
emissions between two similar cars surprising. If the NOX levels for two randomly chosen cars
of this type are independent, find the probability that the difference is at least as large as the
value the attendant observed.
T = C1 - C2
Use this formula
µT = 1.4 - 1.4 = 0
The difference between similar cars should
theoretically be 0.
Remember you can never add STANDARD DEVIATIONS.
σC = 0.3 g/mi so σ2C = (0.3)2 = 0.09
σ2T = σ2C + σ2C
σ2T = 0.09 + 0.09 = 0.18
σT = √ 0.18 = 0.424264 or 0.42 g/mi
Difference found by attendant is 1.9 - 1.1 or 0.8
What is the probability that the difference is 0.8 or more or -0.8 or less?
X:
-1.26
-0.84
-0.42
0.0
T = -0.8
z=
0.42
0.42
0.84
1.26
T = 0.8
= 1.9047
P( T > 0.8 ) = P(z < 1.90) ≈ 1 - 0.9713 ≈ 0.0287
z=
0.42
= -1.9047
P( T < -0.8 ) = P(z < -1.90) ≈ 0.0287
P(T > 0.8) + P(T < -0.8) = 0.0287 + 0.0287 = 0.0574
The probability that the difference is at least as large as the value the attendant observed is
approximately 0.0574 or 5.74%
Dec 6-11:14 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Loser buys the pizza Leona and Fred are friendly competitors in high school. Both are about to take
the ACT college entrance examination. They agree that if one of them scores 5 or more points better
than the other, the loser will buy the winner a pizza. Suppose that in fact Fred and Leona have equal
ability, so that each score varies Normally with mean 24 and standard deviation 2. (The variation is
due to luck in guessing and the accident of the specific questions being familiar to the student.) The
two scores are independent. What is the probability that the scores differ by 5 or more points in either
direction?
Use this formula
F - L > 5 or L - F > 5 (do both ways)
µD = 24 - 24 = 0
Remember you can never add STANDARD DEVIATIONS.
σD = 2 so σ2D = 4
σ2T = σ2D + σ2D
σ2T = 4 + 4
σT = √ 8 = 2.83
X:
-8.49
-5.66
-2.83
0
2.83
z=
2.83
5.66
8.49
T = 5.0
T = -5.0
= -1.7667
P( T < -5 ) = P(z < -1.77) ≈ 0.0384
z=
2.83
= 1.7667
P( T > 5 ) = P(z > 1.77) ≈ 1- 0.9616 ≈ 0.0384
P(T > 5) + P(T < 5) = 0.0384 + 0.0384 = 0.0768
The probability that the difference between their two scores is at least 5 points
is approximately 0.0768 or 7.68%
Dec 6-11:14 PM
Multiple choice: Select the best answer for Exercises 65 and 66, which refer to the following
setting. The number of calories in a 1-ounce serving of a certain breakfast cereal is a random
variable with mean 110 and standard deviation 10. The number of calories in a cup of whole milk
is a random variable with mean 140 and standard deviation 12. For breakfast, you eat 1 ounce of
the cereal with 1/2 cup of whole milk. Let T be the random variable that represents the total
number of calories in this breakfast.
The mean of T is
(a) 110.
(b) 140.
(c) 180.
T = 1C + 0.5M
µT = 110 + 0.5(140) = 180 calories
(d) 195.
(e) 250.
Dec 6-11:14 PM
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Chapter 6 Section 2 Day 4 2016s.notebook
May 05, 2016
Multiple choice: Select the best answer for Exercises 65 and 66, which refer to the following
setting. The number of calories in a 1-ounce serving of a certain breakfast cereal is a random
variable with mean 110 and standard deviation 10. The number of calories in a cup of whole milk
is a random variable with mean 140 and standard deviation 12. For breakfast, you eat 1 ounce of
the cereal with 1/2 cup of whole milk. Let T be the random variable that represents the total
number of calories in this breakfast.
The standard deviation of T is
(a) 22.
(b) 16.
(c) 15.62.
(d) 11.66.
(e) 4.
D
T = 1C + 0.5M
Remember you can never add STANDARD DEVIATIONS.
σC = 10 so σ2C = (10)2 = 100
σM = 12 so σ2M = (12)2 = 144
and
σ2T = σ2C + (0.5)2σ2M
σ2T = 100 + 0.25(144)= 136
σT = √136 = 11.6619 calories
Dec 7-3:48 PM
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