ISE 130 Practice Exam #2 Spring 2026
Instructions: Closed Book
Total Points = 80
Name: ______________________
Points: ______________________
1. True (T) or false (F). (10)
(F) The sample mean is a biased estimator of the population mean.
(T) Exponential distribution is memoryless.
(T) The sampling distribution of the sample mean approaches a normal distribution as
the sample size approaches infinity.
(T) A stem-and-leaf can describe the data distribution graphically.
(F) If X is normally distributed, the ln(X) is lognormally distributed.
2. Consider the following sample of observations on coating thickness for low viscosity
paint: 0.83, 0.88, 0.88, 1.04, 1.09, 1.12, 1.29, 1.31, 1.48, 1.49 (14)
(a) Calculate a point estimate of the population mean coating thickness
(b) Calculate sample median
(c) Calculate lower quartile
(d) Calculate upper quartile
(e) Calculate interquartile range
(f) Calculate sample standard deviation
(g) Calculate an estimate of the standard error of the sample mean
Solution:
∑"#
!$" "! ##.&#
(a) 𝜇̂ = 𝑥̅ = #$ #$
= 1.141
"(&) '"(() #.$)'#.#(
(b) 𝑥* = (
= (
= 1.105
(c) Rank order of q1: k = (n+1)*0.25 = (10+1)*0.25 = 2.75
q1 = 𝑥((.+,) = 𝑥(() + 0.75 ∗ 0𝑥(.) − 𝑥(() 2 = 0.88 + 0.75 ∗ (0.88 − 0.88) = 0.88
(d) Rank order of q3: k = (n+1)*0.75 = (10+1)*0.75 = 8.25
q3 = 𝑥(/.(,) = 𝑥(/) + 0.250𝑥()) − 𝑥(/) 2 = 1.31 + 0.25(1.48 − 1.31) = 1.3525
(e) IQR = q3 - q1=1.3525 - 0.88 = 0.4725
, )
*∑"#
!$" ,! - (""./"))
∑"# )
!$" "! 0 #..,,(,0
(
(f) 𝑆 = #$0#
"#
= #$0#
"#
= 0.054299 → S = 0. 243514
2 $.(&.,#&
(g) 𝜎<"̅ = = = 0.077006
√4 √#$
3. Suppose you add 2 to all observations in a sample. (8)
(a) How does this change the sample mean?
Choose one: 'No change', 'Increases by 2', or 'Two times the old sample mean'.
(b) How does this change the sample standard deviation?
Choose one: 'No change', 'increases by 2', or 'Two times the old standard deviation'.
Instead of adding, now you multiple 2 to all observations in a sample.
(c) How does this change the sample mean?
Choose one: 'No change', 'Increases by 2', or 'Two times the old sample mean'.
(d) How does this change the sample standard deviation?
Choose one: 'No change', 'increases by 2', or 'Two times the old standard deviation'.
Solution:
(a) 'Increases by 2';
(b) 'No change';
(c) 'Two times the old sample mean';
(d) 'Two times the old standard deviation'.
4. A driver's reaction time to a visual stimulus is normally distributed with a mean of 0.4
seconds and a standard deviation of 0.05 seconds. (9)
(a) What is the probability that a reaction requires more than 0.5 seconds?
(b) What is the probability that a reaction requires between 0.4 and 0.5 seconds?
(c) What reaction time is exceeded 90% of the time?
Solution:
$.,0$.&
(a) P(X > 0.5) = 𝑃 ?𝑍 > $.$, B= P(Z > 2) = 1 - 0.97725 = 0.02275
$.&0$.& $.,0$.&
(b) P(0.4 < X < 0.5) = 𝑃 ? $.$,
<𝑍< $.$,
B
= P(0 < Z < 2) = P(Z < 2) - P(Z < 0) = 0.47725
"0$.& "0$.&
(c) P(X > x) = 0.90, then 𝑃 ?𝑍 > $.$,
B = 0.90. Therefore, $.$,
= -1.28 and x = 0.336
Instructions: Closed Book
Total Points = 80
Name: ______________________
Points: ______________________
1. True (T) or false (F). (10)
(F) The sample mean is a biased estimator of the population mean.
(T) Exponential distribution is memoryless.
(T) The sampling distribution of the sample mean approaches a normal distribution as
the sample size approaches infinity.
(T) A stem-and-leaf can describe the data distribution graphically.
(F) If X is normally distributed, the ln(X) is lognormally distributed.
2. Consider the following sample of observations on coating thickness for low viscosity
paint: 0.83, 0.88, 0.88, 1.04, 1.09, 1.12, 1.29, 1.31, 1.48, 1.49 (14)
(a) Calculate a point estimate of the population mean coating thickness
(b) Calculate sample median
(c) Calculate lower quartile
(d) Calculate upper quartile
(e) Calculate interquartile range
(f) Calculate sample standard deviation
(g) Calculate an estimate of the standard error of the sample mean
Solution:
∑"#
!$" "! ##.&#
(a) 𝜇̂ = 𝑥̅ = #$ #$
= 1.141
"(&) '"(() #.$)'#.#(
(b) 𝑥* = (
= (
= 1.105
(c) Rank order of q1: k = (n+1)*0.25 = (10+1)*0.25 = 2.75
q1 = 𝑥((.+,) = 𝑥(() + 0.75 ∗ 0𝑥(.) − 𝑥(() 2 = 0.88 + 0.75 ∗ (0.88 − 0.88) = 0.88
(d) Rank order of q3: k = (n+1)*0.75 = (10+1)*0.75 = 8.25
q3 = 𝑥(/.(,) = 𝑥(/) + 0.250𝑥()) − 𝑥(/) 2 = 1.31 + 0.25(1.48 − 1.31) = 1.3525
(e) IQR = q3 - q1=1.3525 - 0.88 = 0.4725
, )
*∑"#
!$" ,! - (""./"))
∑"# )
!$" "! 0 #..,,(,0
(
(f) 𝑆 = #$0#
"#
= #$0#
"#
= 0.054299 → S = 0. 243514
2 $.(&.,#&
(g) 𝜎<"̅ = = = 0.077006
√4 √#$
3. Suppose you add 2 to all observations in a sample. (8)
(a) How does this change the sample mean?
Choose one: 'No change', 'Increases by 2', or 'Two times the old sample mean'.
(b) How does this change the sample standard deviation?
Choose one: 'No change', 'increases by 2', or 'Two times the old standard deviation'.
Instead of adding, now you multiple 2 to all observations in a sample.
(c) How does this change the sample mean?
Choose one: 'No change', 'Increases by 2', or 'Two times the old sample mean'.
(d) How does this change the sample standard deviation?
Choose one: 'No change', 'increases by 2', or 'Two times the old standard deviation'.
Solution:
(a) 'Increases by 2';
(b) 'No change';
(c) 'Two times the old sample mean';
(d) 'Two times the old standard deviation'.
4. A driver's reaction time to a visual stimulus is normally distributed with a mean of 0.4
seconds and a standard deviation of 0.05 seconds. (9)
(a) What is the probability that a reaction requires more than 0.5 seconds?
(b) What is the probability that a reaction requires between 0.4 and 0.5 seconds?
(c) What reaction time is exceeded 90% of the time?
Solution:
$.,0$.&
(a) P(X > 0.5) = 𝑃 ?𝑍 > $.$, B= P(Z > 2) = 1 - 0.97725 = 0.02275
$.&0$.& $.,0$.&
(b) P(0.4 < X < 0.5) = 𝑃 ? $.$,
<𝑍< $.$,
B
= P(0 < Z < 2) = P(Z < 2) - P(Z < 0) = 0.47725
"0$.& "0$.&
(c) P(X > x) = 0.90, then 𝑃 ?𝑍 > $.$,
B = 0.90. Therefore, $.$,
= -1.28 and x = 0.336
