Download AP Statistics Name: Chapter 7 Worksheet 1. Identify the population

AP Statistics
Chapter 7 Worksheet
Name: _____________________________________
1. Identify the population, the parameter, the sample, and the statistic in each of the following settings.
a) A pediatrician wants to know the 75th percentile for the distribution of heights of 10-year-old boys so she takes a sample of 50
patients and calculates Q3 = 56 inches.
b) A Pew Research Center poll asked 1102 12- to 17-year-olds in the United States if they have a cell phone. Of the respondents, 71%
said yes.
Solution:
a) The population is all 10-year-old boys; the parameter of interest is the 75 th percentile, or Q3. The sample is the 50 10-year-old boys
included in the sample; the statistic is the sample Q3 = 56 inches.
b) The population is all 12- to 17-year-olds in the US; the parameter is the proportion of all 12- to 17-year-olds with cell phones. The
sample is the 1102 12- to 17-year-olds in the sample; the statistic is the sample proportion with a cell phone, p̂ = 0.71.
2. A deck of cards with aces and face cards removed so that only the cards 2 through 10 remain. Suppose you shuffle the deck,
randomly select 5 cards, and note the median value of the cards. For example, if the selected cards were 2, 2, 4, 5, and 9, the median
would be 4. We used Fathom to simulate choosing 500Measures
SRSs of size
5 from
the deck
ofDot
cards.
The graph below shows the
Plot
from
Sample
of Collecti...
distribution of the sample median for these 500 samples.
2
3
4 5 6 7 8 9 10
SampleMedian
a) Is this the sampling distribution of the sample median? Justify your answer.
b) Describe the distribution. Are there any obvious outliers?
c) Suppose that another student prepared a different deck of cards and claimed that it was exactly the same as the one used in the
activity. However, when you took an SRS of size 5, the median was 4. Does this provide convincing evidence that the student’s deck
is different?
Solution:
a) No, since the distribution didn’t use all possible samples of size 5, this isn’t the exact sampling distribution.
b) Shape: The graph is roughly symmetric with a single peak at 6. Center: The mean of the distribution is about 6. Spread: The
values fall mostly between 4 and 8, although there are values as low as 2 and as high as 10. Outliers: There don’t seem to be any
outliers.
c) Getting a sample median of 4 does not provide convincing evidence that the student’s deck is different. Getting a sample median
of 4 or lower occurs fairly often just by chance when taking random samples of size 5 from a deck of cards with the aces and face
cards removed.
3. The superintendent of a large school district wants to know what proportion of middle school students in her district are planning to
attend a four-year college or university. Suppose that 80% of all middle school students in her district are planning to attend a fouryear college or university. What is the probability that a SRS of size 125 will give a result within 7 percentage points of the true
value?
State: We want to find the probability that the proportion of middle school students who plan to attend a four-year college or
university falls between 73% and 87%. That is, P(0.73  p̂  0.87).
Plan:  pˆ  0.80 . Because the school district is large, we can assume that there are more than 10(125) = 1250 middle school students
so  pˆ 
 0.80  0.20 
 0.036 . We can consider the distribution of p̂ to be approximately Normal since np = 125(0.80) = 100  10
125
and n(1 – p) = 125(0.20) = 25  10. DRAW A PICTURE AND SHADE!!!!
Conclude: About 95% of all SRSs of size 125 will give a sample proportion within 7 percentage points of the true proportion of
middle school students who are planning to attend a four-year college or university.
4. Suppose that the number of movies viewed in the last year by high school students has an average of 19.3 with a standard deviation
of 15.8. Suppose we take an SRS of 100 high school students and calculate the mean number of movies viewed by the members of the
sample.
a) What is the mean of the sampling distribution of x ?
b) What is the standard deviation of the sampling distribution of x ? Check whether the 10% condition is satisfied.
Solution:
a)  x  19.3
b)  x 

n

15.8
100
 1.58 . The 10% condition is met because there are more than 10(100) = 1000 high school students.
5. At the P. Nutty Peanut Company, dry roasted, shelled peanuts are placed in jars by a machine. The distribution of weights in the
bottles is approximately Normal, with a mean of 16.1 ounces and a standard deviation of 0.15 ounces.
a) Without doing any calculations, explain which outcome is more likely, randomly selecting a single jar and finding the contents to
weigh less than 16 ounces or randomly selecting 10 jars and finding the average contents to weigh less than 16 ounces.
b) Find the probability of each event described above.
Solution:
a) Since averages are less variable than individual measurements, I would expect the sample mean of 10 jars to be closer, on average,
to the true mean of 16.1 ounces. Thus, it is more likely that a single jar would weigh less than 16 ounces than the average of 10 jars to
be less than 16 ounces.
b) Let x = weight of the contents of a randomly selected jar of peanuts. x is N(16.1, 0.15).
P(x < 16) = normalcdf(lower: –, upper: 16, µ: 16.1, : 0.15) = 0.2525.
0.15 

Let x = average weight of the contents of a random sample of 10 jars. x is N 16.1,
.
10 

DRAW A PICTURE AND SHADE!!!!
0.15 

P( x < 16) = normalcdf  lower:  , upper:16,  : 16.1,  :
 = 0.0175. This answer agrees with the answer to part (a) because this
10 

probability is much smaller than 0.2525.
6. Suppose that the number of texts sent during a typical day by a randomly selected high school student follows a right-skewed
distribution with a mean of 15 and a standard deviation of 35. Assuming that students at your school are typical texters, how likely is
it that a random sample of 50 students will have sent more than a total of 1000 texts in the last 24 hours?
State: What is the probability that the total number of texts in the last 24 hours is greater than 1000 for a random sample of 50 high
school students?
Plan: A total of 1000 texts among 50 students is the same as an average number of texts of
1000
 20 . We want to find P( x > 20),
50
35 

where x = sample mean number of texts. Since n is large (50 > 30), x is approximately N  15,
.
50 

SHADE!!!!
DRAW A PICTURE AND
35 

Do: P( x > 20) ≈ normalcdf  lower:20, upper: ,  :15,  :
 = 0.1562.
50 

Conclude: There is about a 16% chance that a random sample of 50 high school students will send more than 1000 texts in a day.