Answer Key (Second Project)
Contents
1 POST ASSESSMENT (PAGE 1) 3
2 Limits, Continuity and Derivatives (PAGE 2) 7
3 Limits, Continuity and Derivatives (PAGE 3) 9
4 Appendix A 12
List of Figures
1 Table 1 - Normal distribution table for z-scores from -3.49 to 0, with normal
curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2 Table 2 - Normal distribution table for z-scores from 0 to 3.49, with normal
curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1
10. Suppose the mean number of days to germination of a variety of seed is 22, with
standard deviation 2.3 days. Find the probability that the mean germination time of a sam-
ple of 160 seeds will be within 0.5 day of the population mean.
Solution:
mean = x̄ = 22
standard deviation = σ = 2.3
sample size = n = 160
Since the sample size is greater than 30, it will be assumed to be normally distributed.
Therefore, the value of the mean and standard deviation of the sample is,
x̄s = x̄ = 22
σ 2.3
σs = √ = √ ≈ 0.1818
n 160
The formula for finding the z-score becomes,
xs − x̄s
z=
σs
where xs is an actual value within the sample. The probability that the mean of the sample
lies within 0.5 of the population mean is given as,
21.5 − x̄s 22.5 − x̄s
P (21.5 < X̄ < 22.5) = P <z<
σs σs
Substituting known values
21.5 − 22 22.5 − 22
P (21.5 < X̄ < 22.5) = P <z<
0.1818 0.1818
P (21.5 < X̄ < 22.5) = P (−2.75 < z < 2.75)
This means the probability is equal to the area under the normal curve from z = −2.75 to
z = 2.75. Using the tables in Appendix A,we can write
P (21.5 < X̄ < 22.5) = P (−2.75 < z < 2.75)
P (21.5 < X̄ < 22.5) = P (z < 2.75) − P (z < −2.75)
P (21.5 < X̄ < 22.5) = 0.9970 − 0.0030
P (21.5 < X̄ < 22.5) = 0.9940
Therefore, the probability is 0.9940. A
2
1 POST ASSESSMENT (PAGE 1)
Choose the letter of your answer from the given choices. Write
your answers on a separate sheet of paper
1. Suppose the mean cost across the country of a 30 day supply of a generic drug is $46.58
with standard deviation $4.84. Find the probability that the mean of a sample of 100 prices
of a 30 day supplies of this drug will be between $45 and $50.
Solution:
mean = x̄ = 46.58
standard deviation = σ = 4.84
sample size = n = 100
Since the sample size is greater than 30, it will be assumed to be normally distributed.
Therefore, the value of the mean and standard deviation of the sample is,
x̄s = x̄ = 46.58
σ 4.84
σs = √ = √ = 0.484
n 100
The formula for finding the z-score becomes,
xs − x̄s
z=
σs
where xs is an actual value within the sample. The probability that the mean of the sample
lies between 45 and 50 is denoted as,
45 − x̄s 50 − x̄s
P (45 < X̄ < 50) = P <z<
σs σs
Substituting known values
45 − 46.58 50 − 46.58
P (45 < X̄ < 50) = P <z<
0.484 0.484
P (45 < X̄ < 50) = P (−3.26 < z < 7.07)
This means the probability is equal to the area under the normal curve from z = −3.26 to
z = 7.07. Using the tables in Appendix A,we can write
P (45 < X̄ < 50) = P (−3.26 < z < 7.07)
P (45 < X̄ < 50) = P (z < 7.07) − P (z < −3.26)
P (45 < X̄ < 50) = 0.9999 − 0.0006
P (45 < X̄ < 50) = 0.9993
Therefore, the probability is 0.9993. B
3
For No. 2 - 3, scores on a common final exam in a large enroll-
ment, multiple section freshman course are normally distributed
with mean 72.7 and standard deviation 13.1
2. Find the probability that the score XX on a randomly selected exam paper is between 70
and 80.
Solution:
mean = x̄ = 72.7
standard deviation = σ = 13.1
The formula for finding the z-score is,
x − x̄
z=
σ
where x is an actual value within the population. The probability that a random exam score
selected will be between 70 and 80 is denoted by
70 − x̄ 80 − x̄
P (70 < XX < 80) = (P <z<
σ σ
Substituting known values
70 − 72.7 80 − 72.7
P (70 < XX < 80) = P <z<
13.1 13.1
P (70 < XX < 80) = P (−0.21 < z < 0.56)
This means the probability is equal to the area under the normal curve from z = −0.21 to
z = 0.56. Using the tables in Appendix A,we can write
P (70 < XX < 80) = P (−0.21 < z < 0.56)
P (70 < XX < 80) = P (z < 0.56) − P (z < −0.21)
P (70 < XX < 80) = 0.7123 − 0.4168
P (70 < XX < 80) = 0.2955
Therefore, the probability is 0.2955. C
3. Find the probability that the mean score X of 38 randomly selected exam paper is
between 70 and 80.
Solution:
mean = x̄ = 72.7
4
standard deviation = σ = 13.1
sample size = n = 38
Since the sample size is greater than 30, it will be assumed to be normally distributed.
Therefore, the value of the mean and standard deviation of the sample is,
x̄s = x̄ = 72.7
σ 13.1
σs = √ = √ ≈ 2.1251
n 38
The formula for finding the z-score becomes,
xs − x̄s
z=
σs
where xs is an actual value within the sample. The probability that the mean of the sample
lies between 70 and 80 is denoted as,
70 − x̄s 80 − x̄s
P (70 < X̄ < 80) = P <z<
σs σs
Substituting known values
70 − 72.7 80 − 72.7
P (70 < X < 80) = P <z<
2.1251 2.1251
P (70 < X < 80) = P (−1.27 < z < 3.44)
This means the probability is equal to the area under the normal curve from z = −1.27 to
z = 3.44. Using the tables in Appendix A,we can write
P (70 < X < 80) = P (−1.27 < z < 3.44)
P (70 < X < 80) = P (z < 3.44) − P (z < 1.27)
P (70 < X < 80) = 0.9997 − 0.1020
P (70 < X < 80) = 0.8977
Therefore, the probability is 0.8977. A
4. Suppose that in a certain region of the country the mean duration of the first marriages
that end in divorce is 7.8 years, standard deviation 1.2 years. Find the probability that in a
sample of 75 divorces the mean age of the marriages is at most 8 years.
Solution:
mean = x̄ = 7.8
standard deviation = σ = 1.2
sample size = n = 75
5
Since the sample size is greater than 30, it will be assumed to be normally distributed.
Therefore, the value of the mean and standard deviation of the sample is,
x̄s = x̄ = 7.8
σ 1.2
σs = √ = √ ≈ 0.13856
n 75
The formula for finding the z-score becomes,
xs − x̄s
z=
σs
where xs is an actual value within the sample. The probability that the mean of the sample
is less than 8 is,
8 − x̄s
P (8 > X̄) = P >z
σs
Substituting known values
8 − 7.8
P (8 > X̄) = P >z
0.13856
P (8 > X̄) = P (1.44 > z)
This means the probability is equal to the area under the normal curve to the left of z = 1.44.
Using the tables in Appendix A,we can write
P (8 > X̄) = P (1.44 > z)
P (8 > X̄) = 0.9251
Therefore, the probability is 0.9251. D
(*Note: The value for the mean is wrong, it must be 7.8 instead of 7.87, if 7.87 is used,
the answer will not be in the choices, it will be 0.8264)
5. Suppose the mean amount of cholesterol in eggs labeled ”large” is 186 milligrams, with
standard deviation 77 milligrams. Find the probability that the mean amount of cholesterol
in a sample of 144 eggs will be within 22 milligrams of the population mean.
Solution:
mean = x̄ = 186
standard deviation = σ = 77
sample size = n = 144
Since the sample size is greater than 30, it will be assumed to be normally distributed.
Therefore, the value of the mean and standard deviation of the sample is,
x̄s = x̄ = 186
6
σ 77
σs = √ = √ ≈ 6.41667
n 144
The formula for finding the z-score becomes,
xs − x̄s
z=
σs
where xs is an actual value within the sample. The probability that the mean of the sample
lies within 0.5 of the population mean is given as,
164 − x̄s 208 − x̄s
P (164 < X̄ < 208) = P <z<
σs σs
Substituting known values
164 − 186 208 − 186
P (164 < X̄ < 208) = P <z<
6.41667 6.41667
P (164 < X̄ < 208) = P (−3.43 < z < 3.43)
This means the probability is equal to the area under the normal curve from z = −3.43 to
z = 3.43. Using the tables in Appendix A,we can write
P (164 < X̄ < 208) = P (−3.43 < z < 3.43)
P (164 < X̄ < 208) = P (z < 3.43) − P (z < −3.43)
P (164 < X̄ < 208) = 0.9997 − 0.0003
P (164 < X̄ < 208) = 0.9994
Therefore, the probability is 0.9994. C
2 Limits, Continuity and Derivatives (PAGE 2)
Apply the limit laws in evaluating the limit of algebraic functions.
x2 − 16
1. lim
x→4 x − 4
Solution:
x2 − 16 (x − 4)(x + 4)
lim = lim
x→4 x − 4 x→4 x−4
= lim (x + 4)
x→4
= (4) + 4
x2 − 16
lim =8
x→4 x − 4
7
√
2. lim m3 − 3m − 1
m→5
Solution:
√ p
lim m3 − 3m − 1 = (5)3 − 3(5) − 1
m→5
√ √
lim m3 − 3m − 1 = 109
m→5
√
lim m3 − 3m − 1 ≈ 10.44
m→5
t3 − 1
3. lim
t→1 t − 1
Solution:
t3 − 1 (t − 1)(t2 + t + 1)
lim = lim
t→1 t − 1 t→1 t−1
2
= lim(t + t + 1)
t→1
2
=1 +1+1
t3 − 1
lim =3
t→1 t − 1
Use the definition of continuity to determine if the function is con-
tinuous at point x = c. In using table of values, use only 3 values
from the left and from the right with 5 decimal values each.
1
4. f (g) = ; g=3
g−3
Solution:
The function must satisfy these three conditions
(i)f (3) exists
(ii) lim f (g) exists
g→3
(iii) lim f (g) = f (3)
g→3
Testing for condition (i),
1
f (3) =
(3) − 3
1
= undefined, does not exists
0
8
Since the function failed at the first condition, this function is not continuous at g = 3
h2 − 1
5. f (h) = ; h = −1
h+1
Solution:
The function must satisfy these three conditions
(i)f (−1) exists
(ii) lim f (h) exists
h→−1
(iii) lim f (h) = f (−1)
h→−1
Testing for condition (i)
(−1)2 − 1
f (−1) =
(−1) + 1
0
f (−1) = indeterminate, does not exists
0
Since the function failed at the first condition, this function is not continuous at h = −1
2p3 − 16
6. h(p) = ; p=2
p−2
Solution:
The function must satisfy these three conditions
(i)h(2) exists
(ii) lim h(p) exists
p→2
(iii) lim h(p) = h(2)
p→2
Testing for condition (i)
2(2)3 − 16
h(2) =
(2) − 2
0
h(2) = indeterminate, does not exists
0
Since the function failed at the first condition, this function is not continuous at p = 2
3 Limits, Continuity and Derivatives (PAGE 3)
Use the formal definition of derivative f 0 (x) = limh→0 f (x+h)−f
h
(x)
to
determine the slope of the tangent line (or simply, the derivative)
of a given function
7. f (v) = v 2 − v
9
Solution:
(v + h)2 − (v + h) − v 2 + v
f 0 (v) = lim
h→0 h
v + 2vh + h − v − h − v 2 + v
2 2
= lim
h→0 h
2vh + h2 − h
= lim
h→0 h
= lim (2v − 1 + h)
h→0
0
f (v) = 2v − 1
√
8. f (z) = z+1
Solution:
p √
0 (z + h) + 1 − z + 1
f (z) = lim
h→0 h
p √ p √
(z + h) + 1 − z + 1 (z + h) + 1 + z + 1
= lim ·p √
h→0 h (z + h) + 1 + z + 1
z+h+1−z−1
= lim p √
h→0
h (z + h) + 1 + z + 1
h
= lim p √
h→0
h (z + h) + 1 + z + 1
1
= lim p √
h→0
(z + h) + 1 + z + 1
1
=√ √
z+1+ z+1
1
f 0 (z) = √
2 z+1
Apply the differentiation rules in computing the derivatives of an
algebraic, exponential, logarithmic, trigonometric functions and in-
verse trigonometric functions.
3b
9. f (b) =
2b + 1
Solution:
3(2b + 1) − 2(3b)
f 0 (b) =
(2b + 1)2
6b + 3 − 6b
=
(2b + 1)2
10
3
=
(2b + 1)2
10. f (d) = d − 4 csc(d) + sec(d) cot(d)
Solution:
1 cos(d)
f (d) = d − 4 csc(d) + ·
cos(d) sin(d)
1
f (d) = d − 4 csc(d) +
sin(d)
f (d) = d − 4 csc(d) + csc(d)
f (d) = d − 3 csc(d)
f 0 (d) = 1 + 3 csc(d) cot(d)
11
4 Appendix A
Figure 1: Table 1 - Normal distribution table for z-scores from -3.49 to 0, with normal curve
12
Figure 2: Table 2 - Normal distribution table for z-scores from 0 to 3.49, with normal curve
13
