Stat 503
Solutions to Homework #4 (55 points)
[In general each normal distribution question is worth 3 points: 1 for calculating z, 1 for looking up the area correctly, 1
for doing the right thing with it. Sometimes the z and area lookup will have been done earlier, in which case the later
question may be worth fewer points. Solutions are shown using both the table and Excel for easier grading, but only one
of these is required. Sensible interpolation using the table is also allowed for lookups. Points will be deducted for
incorrect statements and for solutions that are difficult to follow. Each number used should be identifiable as to whether is
is a y-value, a z-value, or an area (probability).]
Problem 4.5 (p 131)
In an agricultural experiment, a large uniform field was planted with a single variety of wheat. The
field was divided into many plots (each plot being 700 ft) and the yield (lb.) of grain was measured
for each plot. These plot yields followed approximately a normal distribution with mean 88 lb. and
standard deviation 7 lb. What percentage of the plot yields were
(16)
(a) 80 lb. or more?
(3)
z = (80 – 88) / 7 = -1.14
Using cdf table: Area(–1.14) = 0.1271; Pr(Y > 80) = Pr(Z>–1.14)=1–Pr(Z<–1.14)= 1 – 0.1271 = 0.8729
Using Excel NORMDIST:
Pr(Y > 80) = 1 – NORMDIST(80,88,7,TRUE) = 0.8735
(b) 90 lb. or more?
z = (90 – 88) / 7 = 0.29
Using cdf table: Area(0.29) = 0.6141; Pr(Y > 90) = Pr(Z > 0.29)=1–Pr(Z < 0.29)=1 – 0.6141 = 0.3859
Using Excel NORMDIST:
Pr(Y > 90) = 1 – NORMDIST(90,88,7,TRUE) = 0.3875
(c) 75 lb. or less?
z = (75 – 88) / 7 = -1.86
Using cdf table: Area(–1.86) = 0.0314; Pr(Y < 75) = Pr(Z < –1.86) =0.0314
Using Excel NORMDIST: Pr(Y < 75) = NORMDIST(75,88,7,TRUE) = 0.0316
(d) between 75 and 90 lb.?
(3)
(3)
(2)
z1 = -1 86; z2 = 0.29 (from b. and c.)
Using cdf table: Area(0.29) = 0.6141, Area(–1.86) = 0.0314 from above;
Pr(75 < Y < 90) = Pr(–1.86 < Z < 0.29) = 0.6141 – 0.0314 = 0.5827
Using Excel NORMDIST:
Pr(75 < Y < 90) = NORMDIST(90,88,7,TRUE)-NORMDIST(75,88,7,TRUE) = 0.5808
(e) between 90 and 100 lb.?
z1 = (90 – 88)/7 = 0.29; z2 = (100-88) / 7 = 1.71
Using cdf table: Area(z2) = 0.9564, Area(z1)= 0.6141;
Pr(90 < Y < 100) = Pr(0.29 < Z < 1.71) = 0.9564 – 0.6141 = 0.3423
Using Excel NORMDIST:
Pr(90 < Y < 100) = NORMDIST(100,88,7,TRUE) – NORMDIST(100,88,77,TRUE) = 0.3423
(3)
(f) between 75 and 80 lb.?
(2)
z1 = -1.86; z2 = -1.14 (from a. and c.)
Using cdf table: Area(–1.14) = 0.1271, Area(–1.86) = 0.0314;
Pr(75 < Y < 80) = Pr(–1.86 < Z < –1.14) = 0.1271 – 0.0314 = 0.0957
Using Excel NORMDIST:
Pr(75<Y<80) = NORMDIST(80,88,7,TRUE) –NORMDIST(80,88,7,TRUE) = 0.0949
Problem 4.6 (p 132)
Refer to Exercise 4.5. Let Y represent the yield of a plot chosen at random from the field. Find
(a) Pr(Y > 90)
(4)
(2)
From 4.5 b), Pr(Y > 90) = 0.3859
(b) Pr(75 < Y < 90)
(2)
From 4.5 d), Pr(75 < Y < 90) = 0.5827
Problem 4.9 (p 132)
The serum cholesterol levels of a certain population of boys follow a normal distribution with mean
176 mg/dl and standard deviation 30 mg/dl (as in Example 4.1). What percentage of the boys have
serum cholesterol values
(c) 186 or more?
(18)
(3)
z = (186-176)/30 = 0.33;
Using cdf table: Area(0.33) = 0.6293; Pr(Y > 186) = Pr(Z > 0.33) = 1 – 0.6293 = 0.3707
Using Excel: Pr(Y > 186) = 1 – NORMDIST(186,176,30,TRUE) = 0.3694
(d) 156 or less?
z = (156 – 176) / 30 = –0.67
Using cdf table: Area(–0.67) = 0.2514; Pr(Y < 156) = Pr(Z < –0.67) = 0.2514
Using Excel: Pr(Y < 156) = NORMDIST(156,176,30,TRUE) = 0.2525
(e) 216 or less?
z = (216 – 176) / 30 = 1.33
Using cdf table: Area(1.33) = 0.9082; Pr(Y < 216) = Pr(Z < 1.33) = 0.9082
Using Excel: Pr(Y < 216) = NORMDIST(216,176,30,TRUE) = 0.9088
(f) 121 or more?
z = (121 – 176) / 30 = –1.83
Using cdf table: Area(–1.83) = 0.0336; Pr(Y > 121) = Pr(Z > –1.83) = 1 – 0.0336 = 0.9664
Using Excel: Pr(Y > 121) = 1 – NORMDIST(121,176,30,TRUE) = 0.9666
(g) between 186 and 216?
(3)
(3)
(3)
(2)
From a. and c., Area(0.33) = 0.6293, Area(1.33) = 0.9082;
Pr(186 < Y < 216) = Pr(0.33 < Z < 1.33) = 0.9082 – 0.6293 = 0.2789
or using Excel values, Pr(186 < Y < 216) = 0.9088 – 0.6306 = 0.2782
(h) between 121 and 156?
From b. and d., Area(–1.83) = 0.0336, Area(–0.67) = 0.2514;
Pr(121 < Y < 156) = Pr(–1.83 < Z < –0.67) = 0.2514 – 0.0336 = 0.2178
(2)
or using Excel values, Pr(121 < Y < 156) = 0.2525 – 0.0334 = 0.2191
(i) between 156 and 186?
From a. and b., Area(–0.67)= 0.2514, Area(1.33) = 0.6293;
Pr(156 < Y < 186) = Pr(–0.67 < Z < 1.33) = 0.6293 – 0.2514 = 0.3779
or using Excel values, Pr(156 < Y < 186) = 0.6306 - 0.2525 = 0.3781
Problem 4.11 (pp 132)
For the serum cholesterol distribution of Exercise 4.8, find
(a) the 80th percentile
(2)
(6)
Using cdf table, if Area(z) = 0.7995 (close to 0.8) then z = 0.84.
y =  + z = 176 + 30 × 0.84 = 201.2
The 80th percentile is 201.2.
(3)
(b) the 20th percentile
(3)
Using cdf table, if Area(z) = 0.2005 (close to 0.2) then z = -0.84
y =  + z = 176 + 30 × (-0.84) = 150.8
The 20th percentile is 150.8.
Problem 4.27 (pp 146)
The shell thicknesses of the eggs produced by a large flock of hens follow approximately a normal
distribution with mean equal to 0.38 mm and standard deviation equal to 0.03 mm (as in Example
4.2). Find the 95th percentile of the thickness distribution.
(3)
Using cdf table, if Area(z) = 0.9495 (close to 0.95) then z = 1.64. [May also use z = 1.645 or 1.65.]
y =  + z = 0.38 + 0.03 × 1.64 = 0.429.
The 95th percentile is 0.429 mm.
Problem 4.28 (pp 146)
Refer to the eggshell thickness distribution of Exercise 4.22. Suppose an egg is identified as thinshelled if its shell is 0.32 mm or less.
(c) What percentage of the eggs are thin-shelled?
Find Pr(Y < 0.32). z = (0.32 – 0.38) / 0.03 = – 2.
Using cdf table, Area(–2) = 0.0228. Pr(Y < 0.32) = 0.0228. Therefore, 2.28% of the eggs are thin-shelled.
(d) Suppose a large number of eggs from the flock are randomly packed into boxes of 12 eggs each.
What percentage of the boxes will contain at least one thin-shelled egg? [Hint: First find the
percentage of boxes that will contain no thin-shelled egg.]
The number of thin-shelled eggs in a box follows a binomial distribution with
n = 12 and  = 0.0228. Let Y be the number of thin-shelled eggs.
12 
 (0.0228)0 (0.9772)12 = 0.7582
0
 
Pr(Y = 0) = 
Pr(Y  1) = 1 – Pr(Y = 0) = 1 – 0.7582 = 0.2418.
Therefore, 24.2% of the boxes will contain at least one thin-shelled egg.
(8)
(3)
(5)