STAT 500 Midterm 1 Answers (Penn State University) Latest
Verified Review 2023 Practice Questions and Answers for
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Name:_ SOLUTIONS
Stat 500 Midterm 1 (8 questions) YOU HAVE 3 HOURS TO COMPLETE THIS.
Show all work. Just showing answer without calculations may result in a zero for that
question.
1. A water-heater manufacturer guarantees the electric heating element for a period of
five years. The lifetimes, in months, for a sample of 10 such elements are as follows.
49.3 79.3 186.4 68.4 62.6
65.1 53.2 32.3 40.1 29.3
A. (10 points) Using the methods we learned in class, identify the five number summary
for this set of data.
Ordered data: 29.3 32.3 40.1 49.3 53.2 62.6 65.1 68.4 79.3 186.4
Min = 29.3; Max = 186.4
Median position = (n+1) /2 = 5.5; So, Median = Average of 5th and 6th observation =
(53.2+62.6)/2 = 57.9
Q1 position = (n+1) /4 = 2.75; So, Q1 = 2nd observation + (0.75)*(3rd observation – 2nd
observation) = 32.3 + (0.75)*(40.1 – 32.3) = 38.15
Q3 position =3* (n+1) /4 = 8.25; So, Q1 = 8th observation + (0.72)*(9th observation – 8th
observation) = 68.4 + (0.25)*(79.3 – 68.4) = 71.125
Five Number Summary: (29.3, 38.15, 57.9, 71.13, 186.4)
B. (5 points) Using the boxplot method we learned in class, are there any potential
outliers? If so, what value(s) are they?
IQR = Q3 – Q1 = 71.125 – 38.15 = -11.313
(1.5)*IQR = 49.463
Q1 – 49.463 = 21.662
Q3 + 49.463 = 120.588
1
, Maximum value 186.4 would be a potential outlier.
C. (3 points) Based on the five number summary, what shape would you expect the data
to have? Would you expect the mean to be larger or smaller than the median?
With the outlier, we would expect the shape to be skewed to the right and mean to be
larger than median. But without the outlier, since median is closer to Q3 than Q1, the
distribution is skewed to the left and in that case mean is smaller than the median. (Both
answers are acceptable.)
D. (2 points) What would be the population of interest represented by this data?
The population is all the electric heating elements produced by the water-heater
manufacturer.
2. Past studies have shown that 20% of all surgical patients who receive a particular type of
anesthetic experience mild post-operative nausea. Suppose that five patients are scheduled to
receive this type of anesthetic.
A. (10 points) What is the probability that more than one of the five patients will experience
mild post-operative nausea?
N = 5 and p = 0.2.
To find P(X > 1) = 1 – P(X <=1) = 1 – P(X = 0) – P(X = 1)
P(X = 0) =( 05! 5! ! ) ¿
P(X = 1) = (
1! 4 ! )
5! 1 4
(0.2) ( 0.8) =0.4096
P(X > 1) = = 1 – 0.3277 – 0.4096 = 0.2627
B. (5 points) Of the 5 patients, what is the expected number and standard deviation of patients
that will experience mild post-operative nausea?
Expected number = (5)(0.2) = 1
Standard Deviation = sqrt{(5)(0.2)(0.8)} = 0.894
3. Suppose the body temperatures of adults has a normal distribution with a mean of 98.6° F
and a standard deviation of 0.60° F. 36 adults are randomly selected from the population.
2
Verified Review 2023 Practice Questions and Answers for
Exam Preparation, 100% Correct with Explanations, Highly
Recommended, Download to Score A+
Name:_ SOLUTIONS
Stat 500 Midterm 1 (8 questions) YOU HAVE 3 HOURS TO COMPLETE THIS.
Show all work. Just showing answer without calculations may result in a zero for that
question.
1. A water-heater manufacturer guarantees the electric heating element for a period of
five years. The lifetimes, in months, for a sample of 10 such elements are as follows.
49.3 79.3 186.4 68.4 62.6
65.1 53.2 32.3 40.1 29.3
A. (10 points) Using the methods we learned in class, identify the five number summary
for this set of data.
Ordered data: 29.3 32.3 40.1 49.3 53.2 62.6 65.1 68.4 79.3 186.4
Min = 29.3; Max = 186.4
Median position = (n+1) /2 = 5.5; So, Median = Average of 5th and 6th observation =
(53.2+62.6)/2 = 57.9
Q1 position = (n+1) /4 = 2.75; So, Q1 = 2nd observation + (0.75)*(3rd observation – 2nd
observation) = 32.3 + (0.75)*(40.1 – 32.3) = 38.15
Q3 position =3* (n+1) /4 = 8.25; So, Q1 = 8th observation + (0.72)*(9th observation – 8th
observation) = 68.4 + (0.25)*(79.3 – 68.4) = 71.125
Five Number Summary: (29.3, 38.15, 57.9, 71.13, 186.4)
B. (5 points) Using the boxplot method we learned in class, are there any potential
outliers? If so, what value(s) are they?
IQR = Q3 – Q1 = 71.125 – 38.15 = -11.313
(1.5)*IQR = 49.463
Q1 – 49.463 = 21.662
Q3 + 49.463 = 120.588
1
, Maximum value 186.4 would be a potential outlier.
C. (3 points) Based on the five number summary, what shape would you expect the data
to have? Would you expect the mean to be larger or smaller than the median?
With the outlier, we would expect the shape to be skewed to the right and mean to be
larger than median. But without the outlier, since median is closer to Q3 than Q1, the
distribution is skewed to the left and in that case mean is smaller than the median. (Both
answers are acceptable.)
D. (2 points) What would be the population of interest represented by this data?
The population is all the electric heating elements produced by the water-heater
manufacturer.
2. Past studies have shown that 20% of all surgical patients who receive a particular type of
anesthetic experience mild post-operative nausea. Suppose that five patients are scheduled to
receive this type of anesthetic.
A. (10 points) What is the probability that more than one of the five patients will experience
mild post-operative nausea?
N = 5 and p = 0.2.
To find P(X > 1) = 1 – P(X <=1) = 1 – P(X = 0) – P(X = 1)
P(X = 0) =( 05! 5! ! ) ¿
P(X = 1) = (
1! 4 ! )
5! 1 4
(0.2) ( 0.8) =0.4096
P(X > 1) = = 1 – 0.3277 – 0.4096 = 0.2627
B. (5 points) Of the 5 patients, what is the expected number and standard deviation of patients
that will experience mild post-operative nausea?
Expected number = (5)(0.2) = 1
Standard Deviation = sqrt{(5)(0.2)(0.8)} = 0.894
3. Suppose the body temperatures of adults has a normal distribution with a mean of 98.6° F
and a standard deviation of 0.60° F. 36 adults are randomly selected from the population.
2
