Back to Lesson 11-2
Chapter 11
Normal Distributions
6. Using a statistics utility we find the
probability to be about 0.6450.
Lesson 11-2(pp. 692-697)
1. The mean of the standard normal
distribution is 0 because it has y-axis
symmetry, and the standard deviation is
1, by definition.
2. a. Using the standard normal table, we
find the area below 1.5 to be about
0.9332.
b. The total area under the curve is 1 so
the area to the right of the shaded
area under the curve is given by
1 – 0.9332 = 0.0668.
3. a. The area under the curve below 1.5
is 0.9332, and the area under the
curve below 0 is 0.5, so the area
between 0 and 1.5 is 0.9332 – 0.5 =
0.4332.
b. This is the probability that z is
between -1.5 and 1.5. This is twice
the area between 0 and 1.5 under the
curve. This is 2(0.4332) = 0.8664.
c. The probability that z is greater than
-1.5 is the same (by symmetry) as the
probability that z is less than 1.5,
which is 0.9332.
4. Using a statistics utility we find that the
probability is about 0.9265.
5. Using a statistics utility we find the
probability to be about 0.4761.
315 Functions, Statistics, and Trigonometry Solution Manual
7. Using a statistics utility we find the
probability to be about 0.9987.
8. Using a statistics utility we find the
probability to be about 0.4515.
9. Using the standard normal table we find
the probability that z is less than 2 to be
0.9772. Because of this, we know the
probability that z is greater than 2 is
0.0228. By symmetry, we know the
probability that z is less than -2 is also
0.0228, so the probability that z is
between -2 and 2 is 0.9772 – 0.0228 =
0.9545, or about 95.45%.
10. Using the standard normal table we find
the probability that z is less than 3 to be
0.9987. Because of this, we know the
probability that z is greater than 3 is
0.0013. By symmetry, we know the
probability that z is less than -3 is also
0.0013, so the probability that z is
between -3 and 3 is 0.9987 – 0.0013 =
0.9974, or about 99.7%.
11. The probability that z is less than 1 is
0.8413. The probability that z is less
than -1 is the same as the probability that
z is greater than 1, which is 1 – 0.8413.
This makes the probability that z is
between -1 and 1 equal to 0.8413 –
(1 – 0.8413) = 0.6826.
Chapter 11, Lesson 11-2
Back to Lesson 11-2
12. The values in the table are all above 0.5,
so we will not find 0.1736 on the
standard normal table. By symmetry,
1 – 0.1736 = 0.8264 is on the table,
which will have the opposite sign of the
z-score in question. This corresponds to
0.94. Thus the value we are looking for
is -0.94.
13. The standard normal table shows the
probability that z is less than some
value c. The probability that z is greater
than c is 1 – f(z). Thus, 0.2514 = 1 –
f(z), so f(z) = 0.7486. We can find this
value on the standard normal table; it
corresponds to z = 0.67.
14. The value that is exceeded by 25% so
about 75% will be below the value. This
means we are looking for the value of z
such that f(z) = 0.75. We find that this
corresponds to about 0.67.
15. We know that if 90% of observations
fall within a certain distance of the
mean, that 5% of values must fall above
that distance above the mean. This
means we are looking for the z whose
f(z) is about 0.95. This corresponds to
about 1.645.
20. To find the z-score, subtract the mean
and divide by the standard deviation.
= -1.4 .
This gives 6875
5
21. a. To find the mean, add the values and
divide by the number of values. This
gives x = 122. To find the standard
deviation, subtract the mean from
each value, square each deviation,
add them together, and take the
square root. This gives s = 24.6475.
b. Apply the transformation to each
data point and follow the process
from part a. This gives x = 0 and
s = 1. This makes sense because the
mean of the standard normal curve is
0 and the standard deviation is 1.
22. a. i. 0.5140
ii. 0.5271
b. The graph of the standard normal
curve is nearly a straight line (linear)
on the interval 0 < z < 0.09.
c. The graph of the standard normal
curve is not close to linear in other
areas.
16. We are looking for the percentage of
people who get over 700, which is the
same as the area under the normal curve
above z = 2. This corresponds to
1 – f(2) = 1 – 0.9772 = 0.0228,
or about 2.28%.
17. The area under the parent normal curve
is , while the area under the standard
normal curve is 1.
18. False; we can see by looking at the
parent normal curve that the area under
the parent normal curve is not equal to 1.
19. We know that μ = 40 = np , and
= 4.6 = np(1 p) . Solving these
equations gives n = 85 and p 0.471 .
316 Functions, Statistics, and Trigonometry Solution Manual
Chapter 11, Lesson 11-2