Standard Units and Area Worksheet Answers
1. A normal distribution has 𝜇 = 10 and 𝜎 = 2
a) Find the z – score corresponding to 𝑥 = 12
𝑧=
𝑥 − 𝜇 12 − 10 2
=
= = 1.00
𝜎
2
2
b) Find the z – score corresponding to 𝑥 = 4
𝑧=
𝑥 − 𝜇 4 − 10 −6
=
=
= −3
𝜎
2
2
c) Find the raw score corresponding to 𝑧 = 1.5
𝑥 = 𝑧𝜎 + 𝜇 = 1.5(2) + 10 = 3 + 10 = 13
d) Find the raw score corresponding to 𝑧 = −1.2
𝑥 = 𝑧𝜎 + 𝜇 = −1.2(2) + 10 = −2.4 + 10 = 7.6
2. Use the left tailed z – score table to find each area for each :
Area to the left of 𝑧 = −1.26
area = 0.1038 ( from the table )
Area to the left of 𝑧 = 2.54
area = 0.9945 ( from the table )
Area to the left of 𝑧 = 0.55
area = 0.7088 ( from the table )
Area to the left of 𝑧 = −2.71
area = 0.0034 ( from the table )
Area to the left of 𝑧 = 3.02
area = 0.9987 ( from the table )
Area between 𝑧 = −1.45 and 𝑧 = 0.50
Area = (area z = 0.50 – area z = -1.45 ) = 0.6915 – 0.0735 = 0.618
Area between 𝑧 = 1.6 and 𝑧 = 2.75
Area = (area z = 2.75 – area z = 1.6 ) = 0.9970 – 0.9452 = 0.0518
#2 continued :
Area between 𝑧 = 1.42 and 𝑧 = 2.17
Area = (area z = 2.17 – area z = 1.42) = 0.9850 – 0.9222 = 0.0628
Area to the right of 𝑧 = −1.05
Area = 1 – (area z = -1.05) = 1 – 0.1469 = 0.8531
Area to the right of 𝑧 = 1.95
Area = 1 – (area z = 1.95) = 1 – 0.9744 = 0.0256
3. The college physical education department offered an advanced first aid course last
semester. The scores on the comprehensive final exam were normally distributed, and
some of the z – scores are shown below :
Robert = 1.10
Jacob = 1.70
Susan = - 2.00
Joel = 0.00
Rachel = - 0.80
Lydia = 1.60
a) Who scored above the mean ?
Robert , Jacob , and Lydia
b) Who scored below the mean ?
Susan , and Rachel
c) If the mean score was 𝜇 = 150, with a standard deviation 𝜎 = 20, what was the
final exam score for each student ?
Robert = 𝑧𝜎 + 𝜇 = 1.10(20) + 150 = 22 + 150 = 172
Joel = 𝑧𝜎 + 𝜇 = 0.00(20) + 150 = 0 + 150 = 150
Jacob = 𝑧𝜎 + 𝜇 = 1.70(20) + 150 = 34 + 150 = 184
Rachel = 𝑧𝜎 + 𝜇 = −0.80(20) + 150 = −16 + 150 = 134
Susan = 𝑧𝜎 + 𝜇 = −2.00(20) + 150 = −40 + 150 = 110
Lydia = 𝑧𝜎 + 𝜇 = 1.60(20) + 150 = 32 + 150 = 182
4. Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight
that is approximately normally distributed with mean 𝜇 = 27.2 𝑘𝑔 and standard
deviation 𝜎 = 4.3 𝑘𝑔. Let 𝑥 be the weight of a fawn in kilograms. Convert each of the
following 𝑥 intervals into 𝑧 intervals .
a) 𝑥 < 30
𝑧=
𝑥−𝜇
𝜎
=
30−27.2
4.3
=
2.8
4.3
= 0.65
𝑧 < 0.65
b) 𝑥 > 19
𝑧=
𝑥−𝜇
𝜎
=
19−27.2
4.3
=
−8.2
4.3
= −1.91
𝑧 > −1.91
c) 32 < 𝑥 < 35
𝑧=
𝑥−𝜇
𝜎
=
32−27.2
4.3
=
4.8
4.3
= 1.12
𝑧=
𝑥−𝜇
𝜎
=
35−27.2
4.3
=
7.8
4.3
1.12 < 𝑧 < 1.81
Convert each of the following 𝑧 intervals into 𝑥 intervals
d) 𝑧 > −2.17
𝑥 = 𝑧𝜎 + 𝜇 = −2.17(4.3) + 27.2 = −9.33 + 27.2 = 17.87
𝑥 > 17.87
e) 𝑧 < 1.28
𝑥 = 𝑧𝜎 + 𝜇 = 1.28(4.3) + 27.2 = 5.50 + 27.2 = 32.7
𝑥 < 32.7
f) −1.99 < 𝑧 < 1.44
𝑥 = 𝑧𝜎 + 𝜇 = −1.99(4.3) + 27.2 = −8.56 + 27.2 = 18.64
𝑥 = 𝑧𝜎 + 𝜇 = 1.44(4.3) + 27.2 = 6.19 + 27.2 = 33.39
18.64 < 𝑥 < 33.39
= 1.81
g) What is the probability that a fawn weighs between 22.5 kg and 33.5 kg ?
𝑧=
𝑧=
𝑥 − 𝜇 22.5 − 27.2 −4.7
=
=
= −1.09
𝜎
4.3
4.3
𝑥−𝜇
𝜎
=
33.5−27.2
4.3
=
6.3
4.3
= 1.47
Area = ( area between z = -1.09 and z = 1.47)
Area = ( area z = 1.47) – (area z = - 1.09)
Area = 0.9292 – 0.1379 = 0.7913
5. Let 𝑧 be a random variable with a standard normal distribution. Find the indicated
probability :
a) 𝑃(𝑧 ≤ −0.13) = 0.4483
b) 𝑃(𝑧 ≤ 3.20) = 0.9993
c)
𝑃(𝑧 ≥ −1.50) = 1 − 0.0668 = 0.9332
d) 𝑃(−1.20 ≤ 𝑧 ≤ 2.64) = 0.9959 − 0.1151 = 0.
e) 𝑃(−1.78 ≤ 𝑧 ≤ −1.23) = 0.1093 − 0.0375 = 0.0718
f)
𝑃(0 ≤ 𝑧 ≤ 1.62) = 0.9474 − 0.5000 = 0.4474