Stat 225 - Quiz 5
11/02/2012
Name________________________________________
You have 20 minutes to complete this quiz. Show sufficient work to receive full credit. If a decimal
answer is not exact, please round to 4 non-zero decimal places.
The time Purdue undergraduate students spend watching TV in one week follows a normal distribution
with a mean of 8 hours and a variance of 4 hours2. Strictly use the Empirical Rules to solve the following
problems.
1. What is the probability that a randomly chosen undergraduate student will watch TV less than 14
hours this week? (3 points)
X=the time Purdue undergraduate students spend watching TV in one week
𝑋~𝑁(𝑚𝑒𝑎𝑛 = 8, 𝑆𝐷 = 2)
𝑃 𝑋 < 14 = 𝑃 𝑋 < 𝜇 + 3𝜎 = 1 − .15% = 0.9985
2. What is the probability that a randomly chosen undergraduate student will watch TV between 4 and
14 hours, inclusive, this week. (3 points)
P 4 ≤ X ≤ 14 = P μ − 2σ < 𝑋 < 𝜇 + 3𝜎 = 1 − 0.15% − 2.35% − 0.15% = 0.9735
3. Find the 84th percentile of the time Purdue undergraduate students spend watching TV in one week.
(3 points)
Since P X < 𝜇 + 𝜎 = 84%, 84th percentile is μ + σ = 8 + 2 = 10
Empirical Rule: If 𝑋~𝑁(𝑚𝑒𝑎𝑛 = 𝜇, 𝑆𝐷 = 𝜎), then
𝑃 𝜇 − 𝜎 < 𝑋 < 𝜇 + 𝜎 = 68%; 𝑃 𝜇 − 2𝜎 < 𝑋 < 𝜇 + 2𝜎 = 95%; 𝑃 𝜇 − 3𝜎 < 𝑋 < 𝜇 + 3𝜎 = 99.7%
If Y is a random variable, then 𝑉𝑎𝑟 𝑌 = 𝐸 𝑌 2 − 𝐸[𝑌]2
The cholesterol level for adult males is normally distributed with a mean of 4.8 mmol/L and a standard
deviation of 0.6 mmol/L. Let X denote the cholesterol level for adult males.
4. What is the probability that a randomly chosen adult male has a cholesterol level between 4.7 and 5.5
mmol/L. (3 points)
𝑋~𝑁(𝑚𝑒𝑎𝑛 = 4.8, 𝑆𝐷 = 0.6)
P 4.7 < 𝑋 < 5.5 = P
4.7 − 4.8 X − 4.8 5.5 − 4.8
<
<
= P −0.17 < 𝑍 < 1.17
0.6
0.6
0.6
= P Z < 1.17 − P Z < −0.17 = 0.879 − 0.4325 = 0.4465
5. An adult male is at high risk if his cholesterol level is at least 6.2 mmol/L. What is the probability that a
randomly chosen adult male is at high risk? (3 points)
𝑃 𝑋 ≥ 6.2 = 𝑃
X − 4.8 6.2 − 4.8
≥
= 𝑃 𝑍 ≥ 2.33 = 1 − 𝑃 𝑍 < 2.33
0.6
0.6
= 1 − 0.9901 = 0.0099
6. Compute 𝐸[𝑋 2 ] (4 points)
Var X = E X 2 − E[X]2
so E X 2 = Var X + E[X]2 = (0.6)2 + (4.8)2 = 23.4
𝑋−𝜇
If 𝑋~𝑁(𝑚𝑒𝑎𝑛 = 𝜇, 𝑆𝐷 = 𝜎), then 𝑍 =
and 𝑍~𝑁(𝑚𝑒𝑎𝑛 = 0, 𝑆𝐷 = 1)
𝜎
If X is a continuous random variable with density function f(x), and g(x) is a function of x, then
+∞
𝐸g x
=
𝑔(𝑥) ∙ 𝑓 𝑥 𝑑𝑥
−∞
Stat 225 - Quiz 5
11/02/2012
Name________________________________________
You have 20 minutes to complete this quiz. Show sufficient work to receive full credit. If a decimal
answer is not exact, please round to 4 non-zero decimal places.
The time Purdue undergraduate students spend watching TV in one week follows a normal distribution
with a mean of 12 hours and a variance of 9 hours2. Strictly use the Empirical Rules to solve the
following problems.
1. What is the probability that a randomly chosen undergraduate student will watch TV more than 3
hours this week? (3 points)
X=the time Purdue undergraduate students spend watching TV in one week
𝑋~𝑁(𝑚𝑒𝑎𝑛 = 12, 𝑆𝐷 = 3)
𝑃 𝑋 > 3 = 𝑃 𝑋 > 𝜇 − 3𝜎 = 1 − .15% = 0.9985
2. What is the probability that a randomly chosen undergraduate student will watch TV between 3 and
18 hours, inclusive, this week. (3 points)
P 3 ≤ X ≤ 18 = P μ − 3σ < 𝑋 < 𝜇 + 2σ = 1 − 0.15% − 2.35% − 0.15% = 0.9735
3. Find the 2.5th percentile of the time Purdue undergraduate students spend watching TV in one week.
(3 points)
Since P X < 𝜇 − 2σ = 2.5%, 2.5th percentile is μ − 2σ = 12 − 2 ∙ 3 = 6
Empirical Rule: If 𝑋~𝑁(𝑚𝑒𝑎𝑛 = 𝜇, 𝑆𝐷 = 𝜎), then
𝑃 𝜇 − 𝜎 < 𝑋 < 𝜇 + 𝜎 = 68%; 𝑃 𝜇 − 2𝜎 < 𝑋 < 𝜇 + 2𝜎 = 95%; 𝑃 𝜇 − 3𝜎 < 𝑋 < 𝜇 + 3𝜎 = 99.7%
If Y is a random variable, then 𝑉𝑎𝑟 𝑌 = 𝐸 𝑌 2 − 𝐸[𝑌]2
The cholesterol level for adult males is normally distributed with a mean of 5 mmol/L and a standard
deviation of 1.2 mmol/L. Let X denote the cholesterol level for adult males.
4. What is the probability that a randomly chosen adult male has a cholesterol level between 4.6 and 5.9
mmol/L. (3 points)
𝑋~𝑁(𝑚𝑒𝑎𝑛 = 5, 𝑆𝐷 = 1.2)
P 4.6 < 𝑋 < 5.9 = P
4.6 − 5 X − 5 5.9 − 5
<
<
= P −0.33 < 𝑍 < 0.75
1.2
1.2
1.2
= P Z < 0.75 − P Z < −0.33 = 0.7734 − 0.3707 = 0.4027
5. An adult male is at high risk if his cholesterol level is at least 6.4 mmol/L. What is the probability that a
randomly chosen adult male is at high risk? (3 points)
𝑃 𝑋 ≥ 6.4 = 𝑃
X − 5 6.4 − 5
≥
= 𝑃 𝑍 ≥ 1.17 = 1 − 𝑃 𝑍 < 1.17
1.2
1.2
= 1 − 0.8790 = 0.1210
6. Compute 𝐸[𝑋 2 ] (4 points)
Var X = E X 2 − E[X]2
so E X 2 = Var X + E[X]2 = (1.2)2 + (5)2 = 26.44
𝑋−𝜇
If 𝑋~𝑁(𝑚𝑒𝑎𝑛 = 𝜇, 𝑆𝐷 = 𝜎), then 𝑍 =
and 𝑍~𝑁(𝑚𝑒𝑎𝑛 = 0, 𝑆𝐷 = 1)
𝜎
If X is a continuous random variable with density function f(x), and g(x) is a function of x, then
+∞
𝐸g x
=
𝑔(𝑥) ∙ 𝑓 𝑥 𝑑𝑥
−∞
Stat 225 - Quiz 5
11/02/2012
Name________________________________________
You have 20 minutes to complete this quiz. Show sufficient work to receive full credit. If a decimal
answer is not exact, please round to 4 non-zero decimal places.
The time Purdue undergraduate students spend watching TV in one week follows a normal distribution
with a mean of 11 hours and a variance of 4 hours2. Strictly use the Empirical Rules to solve the
following problems.
1. What is the probability that a randomly chosen undergraduate student will watch TV more than 5
hours this week? (3 points)
X=the time Purdue undergraduate students spend watching TV in one week
𝑋~𝑁(𝑚𝑒𝑎𝑛 = 11, 𝑆𝐷 = 2)
𝑃 𝑋 > 5 = 𝑃 𝑋 > 𝜇 − 3𝜎 = 1 − .15% = 0.9985
2. What is the probability that a randomly chosen undergraduate student will watch TV between 5 and
15 hours, inclusive, this week. (3 points)
P 5 ≤ X ≤ 15 = P μ − 3σ < 𝑋 < 𝜇 + 2σ = 1 − 0.15% − 2.35% − 0.15% = 0.9735
3. Find the 16th percentile of the time Purdue undergraduate students spend watching TV in one week.
(3 points)
Since P X < 𝜇 − σ = 16%, 16th percentile is μ − σ = 11 − 2 = 9
Empirical Rule: If 𝑋~𝑁(𝑚𝑒𝑎𝑛 = 𝜇, 𝑆𝐷 = 𝜎), then
𝑃 𝜇 − 𝜎 < 𝑋 < 𝜇 + 𝜎 = 68%; 𝑃 𝜇 − 2𝜎 < 𝑋 < 𝜇 + 2𝜎 = 95%; 𝑃 𝜇 − 3𝜎 < 𝑋 < 𝜇 + 3𝜎 = 99.7%
If Y is a random variable, then 𝑉𝑎𝑟 𝑌 = 𝐸 𝑌 2 − 𝐸[𝑌]2
The cholesterol level for adult males is normally distributed with a mean of 4.5 mmol/L and a standard
deviation of 1.1 mmol/L. Let X denote the cholesterol level for adult males.
4. What is the probability that a randomly chosen adult male has a cholesterol level between 4.2 and 6.2
mmol/L. (3 points)
𝑋~𝑁(𝑚𝑒𝑎𝑛 = 4.5, 𝑆𝐷 = 1.1)
P 4.2 < 𝑋 < 6.2 = P
4.2 − 4.5 X − 4.5 6.2 − 4.5
<
<
= P −0.27 < 𝑍 < 1.55
1.1
1.1
1.1
= P Z < 1.55 − P Z < −0.27 = 0.9394 − 0.3936 = 0.5458
5. An adult male is at high risk if his cholesterol level is at least 7 mmol/L. What is the probability that a
randomly chosen adult male is at high risk? (3 points)
𝑃 𝑋≥7 =𝑃
X − 4.5 7 − 4.5
≥
= 𝑃 𝑍 ≥ 2.27 = 1 − 𝑃 𝑍 < 2.27
1.1
1.1
= 1 − 0.9884 = 0.0116
6. Compute 𝐸[𝑋 2 ] (4 points)
Var X = E X 2 − E[X]2
so E X 2 = Var X + E[X]2 = (1.1)2 + (4.5)2 = 21.46
𝑋−𝜇
If 𝑋~𝑁(𝑚𝑒𝑎𝑛 = 𝜇, 𝑆𝐷 = 𝜎), then 𝑍 =
and 𝑍~𝑁(𝑚𝑒𝑎𝑛 = 0, 𝑆𝐷 = 1)
𝜎
If X is a continuous random variable with density function f(x), and g(x) is a function of x, then
+∞
𝐸g x
=
𝑔(𝑥) ∙ 𝑓 𝑥 𝑑𝑥
−∞
Stat 225 - Quiz 5
11/02/2012
Name________________________________________
You have 20 minutes to complete this quiz. Show sufficient work to receive full credit. If a decimal
answer is not exact, please round to 4 non-zero decimal places.
The time Purdue undergraduate students spend watching TV in one week follows a normal distribution
with a mean of 17 hours and a variance of 16 hours2. Strictly use the Empirical Rules to solve the
following problems.
1. What is the probability that a randomly chosen undergraduate student will watch TV less than 29
hours this week? (3 points)
X=the time Purdue undergraduate students spend watching TV in one week
𝑋~𝑁(𝑚𝑒𝑎𝑛 = 17, 𝑆𝐷 = 4)
𝑃 𝑋 < 29 = 𝑃 𝑋 < 𝜇 + 3𝜎 = 1 − .15% = 0.9985
2. What is the probability that a randomly chosen undergraduate student will watch TV between 21 and
29 hours, inclusive, this week. (3 points)
P 21 ≤ X ≤ 29 = P μ + σ < 𝑋 < 𝜇 + 3σ = 13.5% + 2.35% = 0.1585
3. Find the 97.5th percentile of the time Purdue undergraduate students spend watching TV in one
week. (3 points)
Since P X < 𝜇 + 2σ = 97.5%, 16th percentile is μ + 2σ = 17 + 2 ∙ 4 = 25
Empirical Rule: If 𝑋~𝑁(𝑚𝑒𝑎𝑛 = 𝜇, 𝑆𝐷 = 𝜎), then
𝑃 𝜇 − 𝜎 < 𝑋 < 𝜇 + 𝜎 = 68%; 𝑃 𝜇 − 2𝜎 < 𝑋 < 𝜇 + 2𝜎 = 95%; 𝑃 𝜇 − 3𝜎 < 𝑋 < 𝜇 + 3𝜎 = 99.7%
If Y is a random variable, then 𝑉𝑎𝑟 𝑌 = 𝐸 𝑌 2 − 𝐸[𝑌]2
The cholesterol level for adult males is normally distributed with a mean of 4 mmol/L and a standard
deviation of 0.8 mmol/L. Let X denote the cholesterol level for adult males.
4. What is the probability that a randomly chosen adult male has a cholesterol level between 4.2 and 5.1
mmol/L. (3 points)
𝑋~𝑁(𝑚𝑒𝑎𝑛 = 4, 𝑆𝐷 = 0.8)
P 4.2 < 𝑋 < 5.1 = P
4.2 − 4 X − 4 5.1 − 4
<
<
= P 0.25 < 𝑍 < 1.38
0.8
0.8
0.8
= P Z < 1.38 − P Z < 0.25 = 0.9162 − 0.5987 = 0.3175
5. An adult male is at high risk if his cholesterol level is at least 6 mmol/L. What is the probability that a
randomly chosen adult male is at high risk? (3 points)
𝑃 𝑋≥6 =𝑃
X−4 6−4
≥
= 𝑃 𝑍 ≥ 2.5 = 1 − 𝑃 𝑍 < 2.5
0.8
0.8
= 1 − 0.9938 = 0.0062
6. Compute 𝐸[𝑋 2 ] (4 points)
Var X = E X 2 − E[X]2
so E X 2 = Var X + E[X]2 = (0.8)2 + (4)2 = 16.64
𝑋−𝜇
If 𝑋~𝑁(𝑚𝑒𝑎𝑛 = 𝜇, 𝑆𝐷 = 𝜎), then 𝑍 =
and 𝑍~𝑁(𝑚𝑒𝑎𝑛 = 0, 𝑆𝐷 = 1)
𝜎
If X is a continuous random variable with density function f(x), and g(x) is a function of x, then
+∞
𝐸g x
=
𝑔(𝑥) ∙ 𝑓 𝑥 𝑑𝑥
−∞