Download Sampling Distributions Worksheet

Sampling Distributions Worksheet – Review Chapter 9
Name_____________
1. Taxi Fares are normally distributed with mean fare $22.27 and a standard deviation of
$2.20.
A) Which should have the greater probability of falling between $21 & $24 – the mean of
a random sample of 10 taxi fares or the amount of a single random taxi fare? Why?
The greater probability of falling between 21 and 24 would be the mean of a random
2.20
sample of 10. The spread in the sample would be smaller: (
) vs 2.20 for the single
10
random fare.
B) Which should have a greater probability of being over $24 – the mean of 10 randomly
selected taxi fares or the amount of a single randomly selected taxi fare? Why?
The chance of a single fare being over 24 is higher than that of a sample of 10. Same
reasons as stated above.
2. Suppose a sample of 50 MP3 players is drawn randomly from a population of MP3
players and the weight, x, of each MP3 player is recorded. Prior experience has shown
that the weight of a single MP3 player has a mean of 6 ounces and a standard deviation of
2.5 ounces.
A) Describe the shape of the sampling distribution of x and justify your answer.
The CLT states that for large samples (n>30) samples will behave approximately normal.
Here n=50.
B) What is the mean and standard deviation of the sampling distribution?
 x  6 oz
x 
2.5
 0.354
50
C) What is the probability that the sample has a mean weight of less than 5 ounces?
Distribution Plot
Normal, Mean=6, StDev=0.354
1.2
1.0
Density
0.8
0.6
0.4
0.002365
0.2
0.0
5
P( x < 5) = 0.002
6
X
normalcdf(-E99, 5, 6, 0.354)
D) How would the sampling distribution of x change if the sample size, n , were
increased from 50 to 100?
Shape is still approximately normal. The mean and standard deviation of the sample
2.5
distribution is  x  6 oz
x 
 0.25 . The standard deviation of the sample
100
distribution of n=100 would be smaller.
3. A soft-drink bottle vendor claims that its process yields bottles with a mean internal
strength of 157 psi (pounds per square inch) and a standard deviation of 3 psi and is
normally distributed. As part of its vendor surveillance, a bottler strikes an agreement
with the vendor that permits the bottler to sample from the vendor’s production to verify
the vendor’s claim.
A) Suppose the bottler randomly selects a single bottle to sample. What is the mean and
standard deviation?
x 157 psi
 x  3 psi
B) What is the probability that the psi of the single bottle is 1.3 psi or more below the
process mean?
Distribution Plot
Normal, Mean=157, StDev=3
0.14
0.12
Density
0.10
0.08
0.06
0.3324
0.04
0.02
0.00
155.7 157
X
normalcdf(-E99, 155.7, 157, 3)
P(x < 155.7) = 0.332
C) Suppose the bottler randomly selected 40 bottles from the last 10,000 produced. What
is the mean and standard deviation of the sampling distribution?
Because the population of bottles is approximately normal, samples will behave
3
approximately normal with  x 157 psi
x 
 0.474 .
40
D) What is the probability that the sample mean of the 40 bottles is 1.3 psi or more below
the process mean?
Distribution Plot
Normal, Mean=157, StDev=0.474
0.9
0.8
0.7
Density
0.6
0.5
0.4
0.3
0.003048
0.2
0.1
0.0
155.7
157
X
normalcdf(-E99, 155.7, 157, 0.474)
P( x < 155.7) = 0.003
E) From part c, in order to reduce the standard deviation 50% (half) , how large would
the sample size need to be.
n = 160. In order to half the standard deviation, n would have to quadruple (times 4). 40
times 4 = 160.
4. A study of 10,000 males who smoke at least two packs of cigarettes daily shows that
the mean life span is 65.3 years while the standard deviation is 3.4 years. If a sample of
40 is selected, find the probability that the mean life span of the sample is less than the
retirement age of 65 years.
The CLT states that samples (n > 30) will behave approximately normal. Here n = 40.
3.4
 x  65.3 years
x 
 0.537
40
Distribution Plot
Normal, Mean=65.3, StDev=0.537
0.8
0.7
0.6
Density
0.5
0.4
0.2882
0.3
0.2
0.1
0.0
P( x < 65) = 0.288
65
65.3
X
normalcdf(-E99, 65, 65.3, 0.537)
5. The U.S. Department of Transportation keeps monthly reports on airline performance.
Among the data which they collect is the percentage of flights that arrive on time. For
November 2006, 78% of the flights by the 12 major airlines arrive on time. For that
period, what is the probability that in a random sample of 100 flights:
Conditions: N > 10n, All flights > 1000
np > 10, 100(.78) > 10, 78 > 10 (check)
n(1-p) > 10, 100(.22) > 10, 22 > 10 (check)
Because the conditions check out, the proportion of flights on time will be approximately
(.78)(.22)
normal with  pˆ  0.78
 pˆ 
 0.041
100
A) More than 63% arrive on time?
Distribution Plot
Normal, Mean=0.78, StDev=0.041
10
Density
8
6
0.9999
4
2
0
0.63
X
0.78
normalcdf(0.63, E99, .78, .041)
P( pˆ > 0. 63) = 0.9999
B) No more than 30% arrive on time.
Distribution Plot
Normal, Mean=0.78, StDev=0.041
10
Density
8
5.8461E-32
6
4
2
0
0.3
P( pˆ < 0. 30) = 0
X
0.78
normalcdf(-E99, .30, .78, 0.041)
6. A battery is designed to last for 25 hours of operation during normal use. The batteries
are produced in batches of 2000 and 50 batteries from each batch are tested. If the mean
life of the sample is less than 24 hours, the entire batch is rejected. Assuming that μ = 25
and σ = 3 hours for any randomly selected battery, find the probability that a batch will
be rejected.
The CLT states that large samples (n > 30) will behave approximately normal. Here n =
50.
3
 x  25 hours
x 
 0.424
50
Distribution Plot
Normal, Mean=25, StDev=0.424
0.9
0.8
Density
0.7
0.6
0.5
0.4
0.009175
0.3
0.2
0.1
0.0
24
25
X
normalcdf(-E99, 24, 25, 0.425)
P( x < 24) = 0.009
7. A sampling distribution for means has a mean of 18 and a standard deviation of 3.2
when the sample size is n = 50. If I need to reduce the standard deviation to ¼ of that,
how large a sample would I need?
In order to reduce the standard deviation by ¼, the sample size needs to increase by a
value 16 times larger.
  22.627
  22.627
3.2 times ¼ = 0.8
x 
 3.2  x 
 0.8
50
800
8. A manufacturing process is designed to produce bolts with a 0.5 inch diameter.. Once a
day, a random sample of 36 bolts is selected and the diameter recorded. If the resulting
sample mean is less than 0.49 inches or greater than 0.51 inches, the process is shut down
for adjustment. The standard deviation for the diameter of a single bolt is 0.02 inches. If
these bolt diameters are normally distributed, what is the probability that the sample
mean falls outside the 0.49 to 0.51 inch interval?
Because the population of bolts is approximately normal, samples will behave
approximately normal.
 x  0.5 in
x 
0.02
 0.003
36
Distribution Plot
Normal, Mean=0.5, StDev=0.003
140
120
100
Density
80
0.9991
60
40
20
0
0.49
0.5
X
0.51
1 – normalcdf(.49, .51, .5, .003)
P(the sample mean falls outside 0.49 to 0.51) = 0.001
9. Just before a referendum on a school budget, a local newspaper polls 400 voters in an
attempt to predict whether the budget will pass. Suppose that the budget actually has the
support of 52% of the voters (like magic we know this…). What’s the probability that the
newspaper’s sample will lead them to predict defeat?
Conditions: N > 10n, All voters > 4,000
np > 10, 400(.52) > 10, 208 > 10 (check)
n(1-p) > 10, 400(.48) > 10, 192 > 10 (check)
Because the conditions check out, the proportion of voters who support the budget will be
(.52)(.48)
 pˆ 
 0.022
approximately normal with  pˆ  0.52
400
Distribution Plot
Normal, Mean=0.52, StDev=0.022
20
Density
15
10
0.1817
5
0
0.5
0.52
X
P( pˆ < 0. 50) = 0.182
normalcdf(-E99, 0.5, 0.52, 0.022)
10. When a truckload of apples arrives at a packing plant, a random sample of 150 is
selected and examined for bruises, discolorations, and other defects. The whole truckload
will be rejected if more than 5% of the sample is unsatisfactory. Suppose that in fact 8%
of the apples on the truck do not meet the desired standard. What’s the probability that
the shipment will be accepted anyway?
Conditions: N > 10n, All apples on the truck > 1,500
np > 10, 150(.08) > 10, 12 > 10 (check)
n(1-p) > 10, 150(.92) > 10, 138 > 10 (check)
Because the conditions check out, the proportion of unsatisfactory apples will be
(.08)(.92)
approximately normal with  pˆ  0.08
 pˆ 
 0.022
150
Distribution Plot
Normal, Mean=0.08, StDev=0.022
20
Density
15
10
0.08634
5
0
0.05
P( pˆ < 0. 05) = 0.086
0.08
X
normalcdf(-E99, 0.05, 0.08, 0.022)