Download HW #6 Solutions

Statistics 371
Homework #6 Solution
Spring 2004
4.2 The 90th percentile of a normal distribution is 1.28 standard deviations above the mean, and
the 10th percentile is 1.28 standard deviations below the mean.
4.5 Plot yields are distributed normally with a mean of 88 lbs and standard deviation of 7 lbs.
Let Y=plot yield.
(a) Pr(Y ≥ 80)=1-Pr(Y<80)=1-Pr(Z <
80−88
7 )=1-Pr(Z
< −1.14)=1-0.1271=0.8729
(b) Pr(Y ≥ 90)=1-Pr(Y<90)=1-Pr(Z <
90−88
7 )=1-Pr(Z
< 0.29)=1-0.6141=0.3859
(c) Pr(Y ≤ 75)=Pr(Z ≤
75−88
7 )=Pr(Z
≤ −1.86)=0.0314
(d) Pr(75 < Y < 90)=Pr(Y < 90)-Pr(Y < 75)=0.6141-0.0314=0.5827
(e) Pr(90 < Y < 100)=Pr(Y < 100)-Pr(Y < 90)=Pr(Z <
100−88
)-Pr(Y
7
< 90)
=Pr(Z < 1.71)-Pr(Y < 90)=0.9564-0.6141=0.3423
(f) Pr(75 < Y < 80)=Pr(Y < 80)-Pr(Y < 75)=0.1271-0.0314=0.0957
4.11 Serum cholesterol levels of 17-year-olds follow a normal distribution with mean 176 mg/dLi
and standard deviation 30 mg/dLi.
Let Y=serum cholesterol level.
(a) To find the 80th percentile of Y:
We want the number ’a’ such that Pr(Y < a)=0.8, or such that Pr(Z <
a−176
30 )=0.8.
From table, Pr(Z < 0.84) ≈ 0.8.
Thus,
a−176
30
= 0.84, and a=201.2
(b) To find the 20th percentile of Y:
We want the number ’a’ such that Pr(Y < a)=0.2, or such that Pr(Z <
a−176
30 )=0.2.
From table, Pr(Z < −0.84) ≈ 0.2.
Thus,
a−176
30
= −0.84, and a=150.8
4. The following gnorm() commands in R (the shaded region in the plots) will show the results
of problem 4.5.
(a) gnorm(88,7,a=80)
(b) gnorm(88,7,a=90)
(c) gnorm(88,7,b=75)
(d) gnorm(88,7,a=75,b=90)
(e) gnorm(88,7,a=90,b=100)
(f) gnorm(88,7,a=75,b=80)
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Statistics 371
Homework #6 Solution
Spring 2004
5. Using qnorm() in R, the 75th quantile of Y ∼ N (2, 4) or Y ∼ N (2, 2 2 ), is found by typing
qnorm(0.75,2,2). Using pnorm() in R, Pr(Y > 2.4), for Y ∼ N (3, 1), is found by typing
1-pnorm(2.4,3,1).
> qnorm(0.75,2,2)
[1] 3.348980
> 1-pnorm(2.4,3,1)
[1] 0.7257469
6(b) The first plot illustrates that the population, or sample space, is non-normal. The second
plot shows that a large sample from this population will approximate the same non-normal
shape. The third plot shows that the sums (or, equivalently, the means), computed from
random samples from this population, WILL be approximately normally distributed.
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