Download Handout on Additional Normal Distribution Exercises

Handout on Additional Normal Distribution Exercises
1. Engineers must consider the widths of heads when designing motorcycle
helmets. It is known that the widths of adult male heads are normally distributed
with mean of 6.1 inches and standard deviation of 1.2 inches.
a. Show a completely labeled normal distribution displaying these population
parameters.
b. Determine the percent of adult males having head widths that are greater
than 8.5 inches.
c. Determine the percent of adult males having head widths that are between
5.5 inches and 6.5 inches.
d. If an adult male is randomly selected from the population, what is the
probability his head width is between 5.5 and 6.5 inches?
2. Women between the ages of 18 and 24 years have systolic blood pressures
that are normally distributed with mean of 114.8 and standard deviation of
13.1.
a. Show a completely labeled normal distribution displaying these population
parameters.
b. What percent of women in this age bracket have systolic blood pressures
that are less than 90?
c. If a doctor wants to identify the 5% of women in this age bracket who have
the highest systolic blood pressures, what pressure defines this group of
women?
3. A study from Consumer Reports showed that the lifetime of TV sets is normally
distributed with mean of 8.2 years and standard deviation of 1.3 years. If is
desired to offer a replacement guarantee on the 2% of sets with the shortest
“lifetimes,” for how long should TV’s be guaranteed?
Answers
1. a.
b. Locate a vertical boundary at x = 8.5 on the distribution as shown below.
Notice that at x = 8.5 the z value is 2. Using the normal distribution
table, A 02 = .4772 and the area to the right of z = 2 is .0228
So, in summary, the percent of males having head widths that are greater
than 8.5 inches is 2.28%
c. Locate vertical boundaries at x = 5.5 inches and at x = 6.5 inches on the
distribution as shown below.
For the boundary on the left
x − μ 5.5 − 6.1 −.6
z=
=
=
= −0.5
σ
1.2
1.2
So, A 0−0.5 = .1915
For the boundary on the right
x − μ 6.5 − 6.1 .4
z=
=
=
≈ 0.33(rounded)
σ
1.2
1.2
So, A 0.33
= .1293
0
.
Adding these two areas will give us the area of the shaded region in the
diagram above: .1915 + .1293 = .3208. So, in summary, the percent of
males having head widths between 5.5 inches and 6.5 inches is 32.08%
d. If one male is selected at random from the population, the answer will be the
same as for part c, namely .3208. In probability notation we could write:
P (5.5 < x < 6.5) = P(−0.5 < z < 0.33) = .3208
2. a.
b. Locate a vertical boundary at the blood pressure 90 on the distribution as
shown below.
At this vertical boundary x = 90 so we can calculate z as follows:
x−μ
90 − 114.8 −24.8
=
≈ − 1.89 (rounded) So, A 0−1.89 = .4706
13.1
13.1
σ
The area of the shaded region is .5000 – .4706 = .0294 or 2.94%. So,
in summary, the percent of females with systolic blood pressures less than
90 is 2.94%.
z=
=
c. The 5% of women who have the highest systolic blood pressures will be in
the right tail of the distribution. The area from the center of the distribution
out to the edge of the region where the highest 5% of blood pressures are
located is 45% or .4500. Thus the z value at this boundary is 1.65.
Since z = 1.65 at the vertical boundary we can now compute the value of x:
x = μ + zσ = 114.8 + (1.65)(13.1) = 114.8 + 21.615 = 136.415 . So, in summary,
the 5% of women having the highest systolic blood pressures will be those
women with blood pressures of about 136 or higher. Note that we round to
the nearest whole number since that is the manner in which blood pressures
are typically reported.
3. Using the mean and standard deviation we can create the following picture:
The 2% of TV’s that will be guaranteed are the TV’s that have the shortest
lifetimes. These will be in the left tail of the distribution, which implies that the
area from the center out to the edge of the guarantee region is 48% = 0.4800 as
shown below.
Using the table for the normal distribution z = –2.05 and then we can write:
x = μ + zσ = 8.2 + (−2.05)(1.3) = 8.2 − 2.665 = 5.535 years.
So, in summary, in order to guarantee 2% of the TV sets, the guarantee
period should be 5.535 years, which converts into about 5 years, 6 months,
and 12 days, or roughly 5½ years.