STAT 571
Qi Jiang
[email protected]
Solution for Homework 6
Problem 1
In this question, we use two R commands: pnorm() and qnorm() to caculate the cumulative probability
and quantile of normal distributed random variable respectively.
1. use R command pnorm()
> pnorm(0.75)
[1] 0.7733726
Using the normal table in the textbook on Page 673, we know P (Z ≥ 0.75) = 0.2663, so P (Z ≤
0.75) = 1 − P (Z ≥ 0.75) = 0.7337
2. use R command pnorm()
> 1-pnorm(-0.97)
[1] 0.8339768
Since standard normal distribution is symmetric P (Z ≥ −0.97) = P (Z ≤ 0.97) = 1 − P (Z ≥ 0.97).
Using the normal table in the textbook on Page 673, we have P (Z ≥ 0.97) = 0.16602, then
P (Z ≥ −0.97) = 1 − P (Z ≥ 0.97) = 0.83398
3. use R command pnorm()
> pnorm(-1.11)-pnorm(-1.18)
[1] 0.01449941
P (−1.18 ≤ Z ≤ −1.11) = P (1.11 ≤ Z ≤ 1.18) = P (Z ≤ 1.18) − P (Z ≤ 1.11) = (1 − P (Z ≥
1.18)) − (1 − P (Z ≥ 1.18))
Using the normal table in the textbook on Page 673, we have P (Z ≥ 1.18) = 0.11900 and P (Z ≥
1.11) = 0.13350.
So, P (−1.18 ≤ Z ≤ −1.11) = (1 − 0.11900) − (1 − 0.1335) = 0.0145
4. Since P ()Z > z) = 0.05 implies that P (Z ≤ z) = 0.95, use R command qnorm()
> qnorm(0.95,0,1)
[1] 1.644854
So z = 1.644854
Using the normal table in the textbook on Page 673, we find the correspondent value that makes
the right tail probability almost equals to 0.05. z is between 1.64 and 1.65.
5. use R command qnorm()
> qnorm(0.1,0,1)
[1] -1.281552
P (Z < z) = 0.10 implies that z is a negative number. However we can only look up positive
numbers in the normal distribution table. Note that P (Z > −z) = 0.1, so we can find −z instead.
Using the normal table in the textbook on Page 673, we find −z is 1.28. Thus z is −1.28.
Office: 1270 MSC
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STAT 571
Qi Jiang
[email protected]
6. P (−z < Z < z) = 0.7 implies that P (Z ≤ z) = 0.85. use R command qnorm()
> qnorm(0.85,0,1)
[1] 1.036433
P (Z ≤ z) = 0.85, so that P (Z > z) = 0.15 Using the normal table in the textbook on Page 673,
we find that z = 1.04.
Problem 2
In the following problems, we use two related R commands:pt() and qt().
1. Use R command
> pt(0.75,10)
[1] 0.764734
In the t distribution table, we know that the 10% right tail quantile is 1.37, which is larger than
0.75. Thus we can make a rough estimation that P (T ≤ 0.75) is less than 0.9.
2. P (T ≥ −0.97) = 1 − P (T < −0.97) Use R command
> 1-pt(-0.97,13)
[1] 0.8251257
Since P (T ≥ −0.97) = P (T ≤ 0.97) and the 10% quantile is 1.35, which is larger than 0.97, so the
rough estimation is less than 0.9.
3. P (−1.18 ≤ T ≤ −1.11) = P (T ≤ −1.11) − P (T ≤ −1.18) Use R command
> pt(-1.11,19)-pt(-1.18,19)
[1] 0.01414329
P (−1.18 ≤ T ≤ −1.11) = P (1.11 ≤ T ≤ 1.18). In the t distribution table, both 1.11 and 1.18 is
between 0(50% quantile) and 10% quantile, so P (−1.18 ≤ T ≤ −1.11) is between 0.5 and 0.9.
4. P (T > z) = 0.05 implies that P (T ≤ z) = 0.95 Use R command
> qt(0.95,11)
[1] 1.795885
So z = 1.795885.
In this case, we find the one tail 5% critical value for a t distribution with df of 11 is 1.80.
5. Use R command
> qt(0.1,15)
[1] -1.340606
So z = −1.340606.
Since P (Z < z) = 0.1 < 0.5 implies z is negative, we would find −z instead. Note P (Z > −z) = 0.1.
In the t table, we find the one tail 10% critical value for a t distribution with a df of 15 is 1.34.
Thus z = −1.34.
Office: 1270 MSC
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STAT 571
Qi Jiang
[email protected]
6. P (−z < Z < z) = 0.7 implies that P (Z ≤ z) = 0.85. Use R command
> qt(0.85,17)
[1] 1.069033
So z = 1.069033.
Also P (−z < Z < z) = 0.7 implies that P (Z > z) = 0.15, but there is no 15% critical value for t
distribution, so we cannot find it in the t table.
Problem 3
We know that X follows a normal distribution with parameters µ = 1.25 and σ = 0.0080. The sample
size n = 25.
1. we want find P (1.26 ≤ X ≤ 1.30 = P (X ≤ 1.30) − P (X < 1.26). Use R,
> pnorm(1.30,1.25,0.0080)-pnorm(1.26,1.25,0.0080)
[1] 0.1056498
Thus P (1.26 ≤ X ≤ 1.30 = 0.1056498.
2. we want find z such that P (X ≥ z) = 0.65. Note P (X ≥ z) = 0.65 implies that P (X < z) = 0.35.
Use R,
> qnorm(0.35,1.25,0.008)
[1] 1.246917
Thus z = 1.246917.
3. Since X is normal distribution N (µ, σ), thus X̄ is N (µ, √σn ) which is N (1.25, 0.0016).
Use R,
>pnorm(1.26,1.25,0.0016)-pnorm(1.24,1.25,0.0016)
[1] 1
4. Use R,
> qnorm(0.9,1.25,0.0016)
[1] 1.25205
So the quantile is 1.25205.
5. Let the cut off values be −z and z. Thus P (X̄ ≤ z) = 0.9.
Use R,
> qnorm(0.9,1.25,0.0016)
[1] 1.25205
Thus the cut off values are −1.25205 and −1.25205.
Office: 1270 MSC
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STAT 571
Qi Jiang
[email protected]
Problem 4
Let X denote the traffic fatalities for each U.S. state resulting from drivers with high alcohol blood levels
in 1982. Then X is normal distributed with parameters of µ = 0.569 and σ = 0.068.
1. we want to find P (X > 0.65).
Use R,
> 1-pnorm(0.65,0.569,0.068)
[1] 0.1167922
So P (X > 0.65) = 0.1167922.
2. > qnorm(0.25,0.569,0.068)
[1] 0.5231347
So the 25th quantile of the fatalities is 0.5231347.
Problem 5
To calculate P (Y > y), we use commands 1 − pnorm(y, µ, σ), and to calculate P (Y < y), we just use
pnorm(y, µ, σ).
1. For µ = 14 and σ = 5, y = 9,
> 1-pnorm(9,14,5)
[1] 0.8413447
> pnorm(9,14,5)
[1] 0.1586553
P (Y > y) = 0.8413447, P (Y < y) = 0.1586553
2. For µ = 15 and σ = 3, y = 18.5,
> 1-pnorm(18.5,15,3)
[1] 0.1216725
> pnorm(18.5,15,3)
[1] 0.8783275
P (Y > y) = 0.1216725, P (Y < y) = 0.8783275
3. For µ = −23 and σ = 4, y = −16,
> 1-pnorm(-16,-23,4)
[1] 0.04005916
> pnorm(-16,-23,4)
[1] 0.9599408
P (Y > y) = 0.04005916, P (Y < y) = 0.9599408.
4. For µ = 14000 and σ = 5000, y = 9000,
Office: 1270 MSC
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STAT 571
Qi Jiang
[email protected]
> 1-pnorm(9000,14000,5000)
[1] 0.8413447
> pnorm(9000,14000,5000)
[1] 0.1586553
P (Y > y) = 0.8413447, P (Y < y) = 0.1586553.
Problem 6
X̄ is distributed as N (µ, √σn ) To calculate P (Y > y), we use commands 1 − pnorm(y, µ, σ), and to
calculate P (Y < y), we just use pnorm(y, µ, σ).
1. For µ = −5 and σ = 5, y = −5.2,
> pnorm(-5.2,-5,5/sqrt(20))
[1] 0.4290138
> pnorm(-5.2,-5,5/sqrt(50))
[1] 0.3886487
#For n=20
#For n=50
Thus P (Ȳ < y) = 0.4290138 when n = 20; P (Ȳ < y) = 0.3886487 when n = 50;
2. For µ = 10 and σ = 30, y = 8,
> pnorm(8,10,30/sqrt(20))
[1] 0.3827972
> pnorm(8,10,30/sqrt(50))
[1] 0.3186759
#For n=20
#For n=50
Thus P (Ȳ < y) = 0.3827972 when n = 20; P (Ȳ < y) = 0.3186759 when n = 50;
3. For µ = −55 and σ = 20, y = −61,
> pnorm(-61,-55,20/sqrt(20))
[1] 0.08985625
> pnorm(-61,-55,20/sqrt(50))
[1] 0.01694743
#For n=20
#For n=50
Thus P (Ȳ < y) = 0.08985625 when n = 20; P (Ȳ < y) = 0.01694743 when n = 50;
4. For µ = 12 and σ = 3, y = 12.5,
> pnorm(12.5,12,3/sqrt(20))
[1] 0.7719717
> pnorm(12.5,12,3/sqrt(50))
[1] 0.8807036
#For n=20
#For n=50
Thus P (Ȳ < y) = 0.7719717 when n = 20; P (Ȳ < y) = 0.8807036 when n = 50;
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