Download Section 7, Normal Dist Practice Exercises answer

PubH 6414 Worksheet 7a: Sampling Distribution of the Mean
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Example 1: Population Distribution
Birth weights at a certain hospital show a normal distribution with mean () of 112 oz. and
standard deviation () of 20.6 oz.
a. What is the probability that the next infant born weighs between 107 and 117 oz? Include
in your answer a sketch of the population distribution and shade the area between 107
and 117 oz.
Area below 107 = 0.4041118
Area below 117 = 0.5958882
Area between 107 and 117 = 0.5958882 - 0.4041118 = 0.1917765
The probability that the next infant born weighs between 107 and 117 oz. is 0.192 (i.e.
19.2% of infants born at this hospital weigh between 107 and 117 oz.)
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 107, 117; mu = 112; sigma = 20.6; select Lower Tail
R script: pnorm(117, mean=112, sd=20.6)-pnorm(107,mean=112,sd=20.6)
Notice that most of the area under the curve is between 50.2 and 173.8.
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b. What is the probability that the next infant born weighs less than 105 oz? Include a sketch
of the population distribution and shade the area less than 105 oz.
Area below 105 = 0.3670014
The probability that the next infant weighs less than 105 ounces is 0.367 (i.e. 36.7% of
infants born at this hospital weigh less than 105 oz.)
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 105; mu = 112; sigma = 20.6; select Lower Tail
R script: pnorm(105, mean=112, sd=20.6)
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Example 2: Birth Weight Sampling Distribution (n= 25)
Suppose we sample 25 infants from the population described in example 1.
a. Draw the sampling distribution of mean birth weights from a sample of 25 infants:
Normal distribution with mean = 112 and SEM = 20.6/5 = 4.12. Label the mean and 1, 2,
3 standard deviations from the mean.
By the CLT the sampling distribution of the sample mean is normally distributed. This
normal distribution has the same mean as the population distribution but a smaller
variance.
b. What is the probability that mean weight for the next 25 infants born is between 107 and
117 oz? Shade the area between 107 and 117 oz.
Area below 107 = 0.1124517
Area below 117 = 0.8875483
Area between 107 and 117 = 0.8875483 - 0.1124517 = 0.7750965
The probability that mean weight for the next 25 infants born is between 107 and 117 oz.
is 0.775.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 107, 117; mu = 112; sigma = 20.6/5; select Lower Tail
R script: pnorm(117, mean=112, sd=20.6/5)-pnorm(107,mean=112,sd=20.6/5)
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The area between 107 and 117 covers a larger proportion of the area under the sampling
distribution of the mean than under the distribution of the individual observations because
the sampling distribution is less dispersed. The SEM is less than the population standard
deviation so a larger proportion of the area under the sampling distribution for samples of
25 is centered near the mean.
c. What is the probability that mean weight for the next 25 infants born is less than 105 oz?
Shade the area less than 105 oz.
Area below 105 = 0.04465685
The probability that mean weight for the next 25 infants born is less than 105 oz is
0.0447.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 105; mu = 112; sigma = 20.6/5; select Lower Tail
R script: pnorm(105, mean=112, sd=20.6/5)
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Example 3: Birth Weight Sampling Distribution (n = 100)
Suppose we sample 100 infants from the population described in example 1.
a. Draw the sampling distribution of mean birthweights from a sample of 100 infants:
Normal distribution with mean = 112 and SEM = 20.6/10 = 2.06. Label the mean and 1,
2, 3 standard deviations from the mean.
b. What is the probability that mean weight for the next 100 infants born is between 107 and
117 oz? Shade the area between 107 and 117 oz.
Area below 107 = 0.007608258
Area below 117 = 0.9923917
Area between 107 and 117 = 0.9923917 - 0.007608258 = 0.9847835
The probability that mean weight for the next 100 infants born is between 107 and 117
oz. is 0.985.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = z_107, z_117; mu = 112; sigma = 20.6/10; select Lower Tail
R script: pnorm(117, mean=112, sd=20.6/10)-pnorm(107,mean=112,sd=20.6/10)
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With larger sample size, the SEM is even smaller – 98.5% of the area under the sampling
distribution of the mean for samples of 100 is between 107 and 117. As sample size
increases the sampling distribution is less dispersed so more of the area under the
sampling distribution is centered near the mean.
c. What is the probability that mean weight for the next 100 infants born is less than 105
oz? Shade the area less than 105 oz.
Area below 105 = 0.0003393298
The probability that mean weight for the next 100 infants born is less than 105 oz.
0.0000339.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 105; mu = 112; sigma = 20.6/10; select Lower Tail
R script: pnorm(105, mean=112, sd=20.6/10)
When sample size increases to 100 only 0.03% of the area under the sampling
distribution is less than 105. It is very unlikely (Probability = 0.000339) that the average
birth weight of 100 infants is less than 105 ounces.
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Example 4: Wing Length Sampling Distribution (n = 64)
Butterfly wing lengths in Baja CA have a skewed distribution with mean () of 4 cm. and
standard deviation () of 5 cm. Suppose we sample 64 butterflies.
a. Draw the sampling distribution of mean wing lengths from a sample of 64:
Normal distribution with mean = 4 cm and SEM = 5/8 = 0.625. Label the mean and 1, 2,
3 standard deviations from the mean
Even if the population distribution isn’t normal, by the CLT this is normal because the
sample size is large enough. Most of the area under the curve is between 2.13 and 5.88.
b. What is the probability that mean wing length from a sample of 64 is between 3.5 and 4.5
cm? Shade the area between 3.5 and 4.5 cm.
Area below 3.5 = 0.2118554
Area below 4.5 = 0.7881446
Area between 3.5 and 4.5 = 0.7881446 - 0.2118554 = 0.5762892
The probability that mean wing length from a sample of 64 is 3.5-4.5 cm is 0.576.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 3.5, 4.5; mu = 4; sigma = 5/8; select Lower Tail
R script: pnorm(4.5, mean=4, sd=5/8)-pnorm(3.5,mean=4,sd=5/8)
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c. What is the probability that mean wing length from a sample of 64 is greater than 5 cm?
Shade the area greater than 5 cm.
Area above 5 = 0.05479929
The probability that mean wing length from a sample of 64 is greater than 5 cm is 0.0548.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 5; mu = 4; sigma = 5/8; select Upper Tail
R script: pnorm(5, mean=4, sd=5/8, lower.tail=FALSE)
1-pnorm(5, mean=4, sd=5/8)
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Example 5: Birth Weight z-transformation of the Sampling Distribution
Birth weights at a certain hospital show a normal distribution with mean () of 112 oz. and
standard deviation () of 20.6 oz. In the following, use the z-transformation of the sampling
distribution of mean birth weights.
a. What is the probability that mean weight for the next 25 infants born is between 107 and
117 oz?
i. Calculate the z-scores (critical ratios) for mean birth weights of 107 and 117 oz.
Then calculate the probability.
z_107 = (107-112)/(20.6/5) = -1.213592
z_117 = (117-112)/(20.6/5) = 1.213592
Area below z_107 = 0.1124517
Area below z_117 = 0.8875483
Area between z_107 and z_117 = 0.8875483 - 0.1124517 = 0.7750965
The probability that mean weight for the next 25 infants born is between 107 and
117 is 0.775.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = z_107, z_117; select Lower Tail
R script: pnorm(z_117)-pnorm(z_107)
ii. The z-transformation of the sampling distribution has mean = 0 and SE = 1.
iii. Compare this probability calculated from the standard normal distribution to the
probability calculated in example 2.2.
The probability calculated from both methods is the same.
b. What is the probability that mean weight for the next 25 infants born is less than 105 oz.
i. Calculate the z-score for mean birth weight of 105 oz. Then calculate the
probability.
z_105 = (105-112)/(20.6/5) = -1.699029
Area below z_105 = 0.04465685
The probability that mean weight for the next 25 infants born is less than 105 oz.
is 0.0447.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = z_105; select Lower Tail
R script: pnorm(z_105)
ii. Compare this probability with the probability calculated in example 2.3.
The probabilities are the same.
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c. What are the two birth weights such that 95% of mean birth weights from samples of size
25 are between these two weights?
i. Report the critical ratios corresponding to this area.
-1.96, 1.96
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Quantiles
Probabilities = 0.025, 0.975; select Lower Tail
R script: qnorm(c(0.025,0.975))
ii. Draw a standard normal distribution and shade in the middle 95% of the area.
What % is in each tail of the distribution?
There is 2.5% in each tail.
iii. What two birth weights correspond to this area?
x1  112
x 2  112
 1.96
 1.96
4.12
4.12
x1  1.96 * 4.12  112
x 2  1.96 * 4.12  112
x1  103.9
x 2  120.1
95% of mean birth weights from samples of 25 are between 103.9 oz and 120.1
oz.
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d. What are the two birth weights such that 95% of mean birth weights from samples of 100
are between these two weights? Include the critical ratios used in your answer.
Critical ratios = ±1.96
x1  112
 1.96
2.06
x1  1.96 * 2.06  112
x 2  112
 1.96
2.06
x 2  1.96 * 2.06  112
x1  107.9
x 2  116.04
95% of mean birth weights from samples of 100 are between 107.9 oz and 116.04 oz.
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Example 6: Wing Length z-transformation of the Sampling Distribution
Butterfly wing lengths in Baja CA have a skewed distribution with mean () of 4 cm. and
standard deviation () of 5 cm. In the following, use the z-transformation of the sampling
distribution of mean wing lengths.
a. What is the probability that mean wing length in a sample of 64 is between 3.5 and 4.5
cm? Include in your answer, the z-scores (critical ratios) and a comparison of the
probability calculated here to that in example 4c.
z_3.5 = (3.5-4)/(5/8) = -0.8
z_4.5 = (4.5-4)/(5/8) = 0.8
Area below z_3.5 = 0.2118554
Area below z_4.5 = 0.7881446
Area between z_3.5 and z_4.5 = 0.7881446 - 0.2118554 = 0.5762892
The probability that mean wing length in a sample of 64 is 3.5-4.5 cm is 0.562.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = z_3.5, z_4.5; select Lower Tail
R script: pnorm(z_4.5)-pnorm(z_3.5)
b. What is the probability that mean wing length in a sample of 64 is greater than 5 cm?
Include in your answer, the z-score (critical ratio) and a comparison of the probability
calculated here to that in example 4e.
z_5 = (5-4)/(5/8) = 1.6
Area below z_5 = 0.05479929
The probability that mean wing lent in a sample of 64 is greater than 5 cm is 0.0548.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = z_5; select Upper Tail
R script: pnorm(z_5, lower.tail=FALSE)
1-pnorm(z_5)
c. Find the two mean wing lengths such that 95% of wing lengths from samples of 64 are
between these two lengths. Include the critical ratios used in your answer.
Critical ratios : ±1.96
The central 95% of the area under the standard normal curve is between -1.96 and 1.96.
x1  4
 1.96
0.625
x1  1.96 * 0.625  4
x1  2.775
x2  4
 1.96
0.625
x 2  1.96 * 0.625  4
x 2  5.225
95% of mean wing lengths from samples of 64 are between 2.775 and 5.225 cm.
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Example 7: Cholesterol Sampling Distribution
Cholesterol levels for women age 40-50 are approximately normally distributed with mean () of
180 mg/dl and standard deviation () of 40 mg/dl. Use the methods in previous exercises to
answer the following parts 1-8.
a. Describe the distribution of the sample mean cholesterol for samples of size n = 16.
Normally distributed with mean 180 and SEM 40/4 = 10.
b. What is the probability that the sample mean cholesterol from a sample of size n = 16 is
between 170 and 190?
Area below 190 = 0.8413447
Area below 170 = 0.1586553
Area between 170 and 190 = 0.8413447- 0.1586553 = 0.6826895
Notice that 170 = mean – 1*SEM and 190 = mean + 1*SEM. The area within 1 standard
error of the mean is 0.68.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 190,170; mu=180; sigma=40/4, select Lower Tail
R script: pnorm(190,mean=180,sd=40/4)-pnorm(170,mean=180,sd=40/4)
c. What is the probability that the sample mean cholesterol from a sample of size n= 16 is >
210?
Area above 210 = 0.001349898
The probability that thesample mean cholesterol from a sample of size 16 is greater than
220 mg/dL is 0.00135.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 210; mu=180; sigma=40/4, select Upper Tail
R script: pnorm(210,mean=180,sd=40/4, lower.tail=FALSE)
1- pnorm (210,mean=180,sd=40/4)
d. 95% of mean cholesterol values for samples of size 16 are between 160.4 and 199.6.
Use the z-scores for the middle 95% (-1.96 and 1.96) of the area under the standard
normal distribution and solve for the mean cholesterol values
x1  180
x 2  180
 1.96
 1.96
10
10
x1  1.96 *10  180
x 2  1.96 *10  180
x1  160.4
x 2  199.6
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e. Describe the distribution of sample mean cholesterol for samples of size n=64.
Normally distributed with mean 180 and SEM 40/8 = 5.
f. What is the probability that the sample mean cholesterol from a sample of size n = 64 is
between 170 and 190?
Area below 170 = 0.02275013
Area below 190 = 0.9772499
Area between 170 and 190 = 0.9772499 - 0.02275013 = 0.9544997
The probability that the sample mean cholesterol from a sample of size n=64 is between
170 and 190 is 0.954.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 190,170; mu=180; sigma=40/8, select Lower Tail
R script: pnorm(190,mean=180,sd=40/8)-pnorm(170,mean=180,sd=40/8)
g. What is the probability that the sample mean cholesterol from a sample of size n= 64 is >
210?
Area above 210 = 9.865876e-10 = 0.000000000986
Therefore, the probability that the sample mean cholesterol from a sample size of n=64 is
greater than 210 mg/dL is extremely unlikely.
Rcmdr: Distributions > Continuous Distributions > Normal Distribution > Normal
Probabilities
Variable value = 210; mu=180; sigma=40/8, select Upper Tail
R script: pnorm(210,mean=180,sd=40/8, lower.tail=FALSE)
1- pnorm (210,mean=180,sd=40/8)
h. 95% of mean cholesterol values for samples of size 64 are between 170.2 and 189.8.
x1  180
 1.96
5
x1  1.96 * 5  180
x 2  180
 1.96
5
x 2  1.96 * 5  180
x1  170.2
x 2  189.8
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Sampling distributions of the mean can be confusing because the distribution of all possible
sample means of a given sample size is an abstract concept. We don’t observe all possible
sample means but, by the CLT, we can describe how these unobserved sample means are
distributed.
The distribution of sample means (x-bars) has a mean () that is equal to the mean () of the
original population distribution. This makes sense if you think about taking a random sample
from the population. Typically, some of the observations in a sample will be greater than the
population mean and some will be less than the population mean so the sample means (x-bars)
will tend to be closer to the population mean () than the original observations. Some, but not
all, sample means will be exactly equal to the population mean. This is why there’s a
distribution of sample means (x-bars) with some less than the population mean () and some
greater than the population mean (), and some equal to the population mean ().
The distribution of sample means (x-bars) has a standard error of the mean (SEM) that is equal to
the population standard deviation () divided by the square root of the sample size. As sample
size increases, the SEM is smaller because the square root of the sample size is in the
denominator of the calculation of SEM. A smaller SEM results in a sampling distribution that is
less dispersed, that is, the sample means are more tightly clustered around the population mean
().
If you look back over these practice exercises, you’ll see the pattern of less dispersed sampling
distributions of the mean with increasing sample size. A less dispersed sampling distribution of
the mean has
 Greater probability of sample means near the center of the sampling distribution
 Smaller probability of sample means in the tails of the sampling distribution