Download 06_Normal Distributions and Sampling Distributions SSS Answers

Normal Distributions and Sampling Distributions
We have intentionally included more material than can be covered in most Student Study
Sessions to account for groups that are able to answer the questions at a faster rate. Use your
own judgment, based on the group of students, to determine the order and selection of questions
to work in the session. Be sure to include a variety of types of questions (multiple choice and free
response) in the time allotted.
Multiple Choice
1. D (Similar to 2007 Q22)
A male grey kangaroo with height 2.05 meters has a z-score of z 
This would correspond to 1.2 
2.05  1.75 0.3

 1.2 .
0.25
0.25
x  1.2
; x  1.2  0.12  1.32 meters for a male red kangaroo.
0.1
2. C (AP Style Question)
P(65  height  90)  P(height  90)  P(height  65)  0.7881  0.1151  0.6731
3. E (AP Style Question)
Since the sample size is greater than thirty it is reasonable (by the CLT) to assume the
sampling distribution will be normally distributed. The mean is equal to the population mean

4
and the standard deviation will be  x  x 
 0.632
n
40
4. C (AP Style Question)


.35  .45
P ( pˆ  .35)  P  z 

(.45)(.55)

100



  P ( z  2.01)  0.022



5. D (2002 Q10)
4000
x  10
x  10
 0.4 ; 0.52 
; 0.52 
; 9.895 cm to 10.105 cm since this interval
10000
0.2
0.2
represents the middle 40% of the distribution
6. B (AP Style Question)



12.2  12 
P( x  12.2)  P  z 
  P( z  2)  0.023
0.2 



4 

7. D (Similar to 2007 Q3)
P ( z  z*)  0.25 therefore z*  0.6745 ; Solve 0.6749 
8. E (Similar to 2002 Q8)


100 
0.33  0.28

z 
P  pˆ 

P

300 

(0.28)(0.72)


300

x 5
; x  5.084 centimeters
0.5
16






9. B (Similar to 2007 Q1)
The interval 45 to 49 inches corresponds to x  1s so approximately 68% (or 272) of the
fourth graders heights should fall between those two values.
10. B (AP Style Question)
z  2.33 corresponds to the 99th percentile; 2.33 
deviation for production should be   0.52 yards
291.2  290

therefore the standard
Free Response:
11. Similar to 2006B Q3
Solution
Part (a):
Let R represent the height a randomly selected ball rebounds to. Since R is normally
distributed with a mean of 50 inches and standard deviation of 1.1 inches,
48  50 
53  50 


P( R  48 or R  53)  P  z 
  P z 

1.1 
1.1 


 P  z  1.8182   P  z  2.727 
 0.0345  0.0032
 0.0377
About 3.7% of the balls produced would not meet the tournament specifications for rebound.
Part (b):
Let X be the number of balls that fail to meet rebound specifications out of 9 randomly
selected balls.
P( X  1)  P( X  0)  P( X  1)
9
9
   (0.0377)0 (0.9623)9    (0.0377)1 (0.9623)8
0
1
 0.7075  0.2496
 0.9571
Part (c):
Since the distribution of rebound heights is normally distributed, which is a mound shaped
symmetric distribution, the maximum area (proportion) between two values would occur
when the mean of the distribution was equal to the mean of the two values. Since the mean of
48 and 53 is 50.5 inches, the manufacturer should set the production so the mean rebound
height is 50.5 inches.
Part (d):
Using the mean determined from part (c), the maximum proportion of balls this manufacturer
produces that meet the high altitude specifications for rebound would be
P(48  R  53)  P( R  53)  P ( R  48)
53  50.5 
48  50.5 


 P z 
  P z 

1.1 
1.1 


 P  z  2.27   P  z  2.27   0.9884  0.0116  0.9768
Scoring:
All parts are scored as essentially correct (E), partially correct (P), or incorrect (I).
Part (a) is essentially correct (E) if the student clearly shows ALL three of the following:
Indicates the distribution is normal.
Specifies BOTH the mean,  , and the standard deviation,  .
Calculates the correct probability.
Part (a) is partially correct (P) if the student:
Calculates the correct probability but fails to specify BOTH  and  in the identification of
the normal distribution.
OR
Completely identifies the distribution as normal with BOTH  and  correctly specified
but fails to calculate the correct probability, e.g., calculates the probability a ball is rebounds
to a height between 48 and 53 inches.
OR
Completely identifies the distribution as normal with BOTH  and  correctly specified
but uses the empirical rule to provide an approximate answer.
Part (a) is incorrect (I) if the student:
Reports a correct probability without showing any work.
OR
Calculates an incorrect probability using an inappropriate distribution.
Notes:
 Calculator solution is 0.0377. If this is the only information provided, the response is
scored as incorrect (I).
 If only the calculator command 1  normalcdf (48, 53, 50, 1.1) is provided along 0.0377
then the response should be scored as partially correct (P).
 If the calculator command 1  normalcdf (48, 53, 50, 1.1) is provided along 0.0377 AND
the mean and standard deviation are clearly identified, then the response should be scored
as essentially correct (E).
 If the calculator command 1  normalcdf (48, 53, 50, 1.1) AND a shaded/labeled sketch
of an appropriate normal distribution are provided along with 0.0377, then the response
should be scored as essentially correct (E).
 Minor arithmetic or transcription errors will not necessarily lower the score.
Part (b) is essentially correct (E) if the student calculates the correct probability and:
Clearly indicates the distribution is binomial AND specifies both n and p using the value
obtained in part (a)
Part (b) is partially correct (P) if the student:
Clearly indicates the distribution is binomial AND specifies both n and p using the value
obtained in part (a), but does not calculate the probability correctly
OR
Calculates the correct probability using the value obtained in part (a) but fails to completely
identify the distribution as binomial with both n and p specified.
OR
Indicates a correct procedure for computing the probability but uses a value of p but uses a
value of p that is different from the value obtained in part (a).
Part (b) is incorrect (I) if the student:
Provides a probability, but no work is shown.
OR
Obtains a probability with an incorrect solution strategy.
Part (c) is essentially correct (E) if the student clearly states BOTH of the following:
 The distribution is normally distributed and therefore symmetric and mound shaped
 The maximum area contained between two values occurs when the mean of the two
values is equal to the mean of the distribution, the mean should be placed at the mean of
the boundary values, 50.5 inches
Part (c) is partially correct (P) if the student:
States the distribution should be centered at 50.5 inches but gives a weak justification
OR
States that the distribution should be centered between 48 and 53 inches but fails to give the
value 50.5 inches.
Part (c) is incorrect (I) if the student:
Gives only the numerical response 50.5 inches with no justification
OR
Gives an incorrect value for the center between 48 and 53
Part (d) is essentially correct (E) if the student clearly shows BOTH of the following:
 Identifies the mean and standard deviation of the distribution using the values indicated
in part (c) and the problem stem respectively,   50.5 and   1.1
 Calculates the correct probability using the values indicated in part (c) and the problem
stem respectively.
Part (d) is partially correct (P) if the student:
Computes the correct probability but fails to indicate the mean used from part (c)
Part (d) is incorrect (I) otherwise.
Each essentially correct (E) response counts as 1 point, each partially correct (P) response counts
1
as point.
2
4
Complete Response
3
Substantial Response
2
Developing Response
1
Minimal Response
1
points), use a holistic approach to
2
determine whether to score up or down depending on the strength of the response and
communication.
Note: If a response is in between two scores (for example, 2
12. Similar to 2004B Q3
Solution
Let X  volume of crude oil in a randomly selected barrel.
Part (a):
55.4  55 

P( X  55.4)  P  z 
  P ( z  0.8)  0.2119
0.5 

Part (b):
No. Approximately 21% of the barrels will have contents of 55.4 gallons or greater when the
filling equipment is working properly, so a barrel that was filled with 55.4 gallons of oil would
not be an unusual occurrence.
Part (c):



55.4  55 
P( X  55.4)  P  z 
  P( z  3.10)  0.0010
0.5 



15 

Part (d):
Yes, we would suspect that the filling mechanism is overfilling. If it is working properly, the
probability that the mean volume of oil in 15 randomly selected cars is 55.4 gallons or greater is
0.0010 , which is very small.
Note: To receive complete credit for part (a) or part (c) students must show how the probability
is computed. Since part (a) and part (c) involve different normal distributions, it is important to
identify which normal distribution is used in each part. As shown above, this could be done by
displaying a probability statement containing the mean and standard deviation for the appropriate
normal distribution. It could be done in other ways, such as listing the mean and standard
deviation and displaying an appropriate graph.
Scoring
Parts (a) and (b) are scored together as section 1 and parts (c) and (d) are scored together as
section 2. Section 1 and 2 are each scored as either essentially correct, partially correct, or
incorrect.
Section 1 is essentially correct if
1. the probability in part (a) is correctly computed (except for minor arithmetic errors as
long as the given answer is a number between 0 and 0.5)
2. the response in part (b) says it would not necessarily indicate that the filling equipment
was malfunctioning and the justification is based on the probability computed in part (a).
Section 1 is partially correct if the probability computed in part (a) is not correct, but the
conclusion in (b) is reasonable relative to the computed probability and the justification is based
on the computed probability,
OR
the probability in part (a) is correctly computed, except for minor arithmetic errors, but the
justification in part (b) is not linked to the computed probability,
OR
the probability in part (a) is correctly computed, except for minor arithmetic errors, and the
conclusion in part (b) is not consistent with the computed probability.
Section 2 is essentially correct if
1. the probability in part (c) is correctly computed (except for minor arithmetic errors as
long as the given answer is a number between 0 and 0.5) using the mean and standard
deviation of the X distribution
2. the response in part (d) says we would suspect that the filling mechanism was overfilling
3. the justification is based on the probability computed in part (c).
Section 2 is partially correct if
the probability computed in part (c) is not correct, but the conclusion in (d) is reasonable relative
to the computed probability and the justification is based on the computed probability
OR
the probability in part (c) is correctly computed except for minor arithmetic errors, but the
justification in part (d) is not linked to the computed probability.
Note: The response in part (b) could be justified by indicating that the 55.4 gallons is less than
one standard deviation away from desired mean of 55 gallons. The response in part (d) could be
justified by indicating that the 55.4 gallons is more than two standard deviations above the
desired mean of 55 gallons.
4
Complete Response
Both sections are essentially correct
3
Substantial Response
One section is essentially correct and the other section is partially correct
2
Developing Response
One section is essentially correct and the other section is incorrect
OR
Both sections are partially correct
1
Minimal Response
One section is partially correct
13. 2010 Q2
Intent of Question
The primary goals of this question were to assess students’ ability to (1) describe a sampling
distribution of a sample mean; (2) set up and perform a normal probability calculation based on
the sampling distribution.
Solution
Part (a):
The sampling distribution of the sample mean song length has mean  X    3.9 minutes

1.1
and standard deviation  X 

 0.174 minutes. The central limit theorem (CLT)
n
40
applies in this case because the sample size ( n  40 ) is fairly large, especially with the
population of song lengths having a roughly symmetric distribution. Thus, the sampling
distribution of the sample mean song length is approximately normal.
Part (b):
The probability that the total airtime of 40 randomly selected songs exceeds the available
time (that is, the probability that the total airtime of 40 randomly selected songs is greater
than 160 minutes) is equivalent to the probability that the sample mean length of the 40 songs
160
 4.0 minutes.
is greater than
40
According to part (a), the distribution of the sample mean length of X is approximately
normal. Therefore,
4.0  3.9 

P( X  4.0)  P  Z 
  P( Z  0.57)  1  0.7157  0.2843 .
0.174 

(The calculator gives the answer as 0.2827.)
The approximate sampling distribution of the sample mean song length and the desired
probability are displayed below.
Part (b) (alternative):
An equivalent approach is to note that the sampling distribution of the total airtime, T, for the
40 songs is approximately normal, with mean 40(3.9)  156 minutes and standard deviation
40(1.1)  6.96 minutes. The z-score for a total airtime of 160 minutes is then
160  156
z
 0.57 , and the calculation proceeds as above.
6.96
Scoring
Parts (a) and (b) are scored as essentially correct (E), partially correct (P) or incorrect (I).
Part (a) is scored as follows:
Essentially correct (E) if the student correctly provides all three components of the sampling
distribution: shape (approximately normal), center (mean 3.9) and spread (standard deviation
1.1
 0.174 ).
40
Partially correct (P) if the student correctly provides only two of the three components.
Incorrect (I) if the student correctly provides only one or none of the components.
Notes
 Describing the sampling distribution as normal instead of approximately normal does not
earn credit for the shape component.
 To earn credit for the spread component, the response must show how the standard
deviation is calculated.
 If a response contains incorrect notation or terminology, it can at best be scored as
partially correct (P).
Part (b) is scored as follows:
Essentially correct (E) if the student sets up and performs a correct normal probability
calculation.
Partially correct (P) if the student sets up the normal probability calculation correctly but
does not carry it through correctly OR sets up an incorrect but plausible calculation (for
example, by using an incorrect standard deviation) but carries it through correctly.
Incorrect (I) if the student does not set up or perform the normal probability calculation
correctly.
Notes
 A student can earn a score of essentially correct (E) in part (b) even with incorrect
parameter values in part (a) by providing a correct calculation that uses the mean and
standard deviation from part (a).
 Calculator syntax: An answer containing “normalcdf(…)” with no additional work or
labeling is a best partially correct (P). If an appropriate sketch with the mean and standard
deviation correctly labeled accompanies the calculator command, OR if the mean and
standard deviation used in the calculator command are clearly identified in part (a) or part
(b), then the response should be scored as essentially correct (E).
 If the student uses the sampling distribution of the total amount of time, T, needed to play
40 randomly selected songs to do the probability calculation, the student must show how
the standard deviation is calculated – unless this value is carried forward from part (a) –
for the response to be scored as essentially correct (E). For example,
 T  40 X  40(1.1)  6.96
4
OR
 T  40 X  40(0.174)  6.96
Complete Response
Both parts essentially correct
3
Substantial Response
One part essentially correct and one part partially correct
2
Developing Response
One part essentially correct and one part incorrect
OR
Both parts partially correct
1
Minimal Response
One part partially correct and one part incorrect