STAT 113 Week 9 (Mar 4 )
Work Sheet 6: Chapter 13
 Normal Curve
1. The following figure shows 4 different normal curves.
(a) Rank the curves in the increasing order of means. (Hint: some of the graphs may
have equal means.)
c = d < a <b
(b) Rank the curves in the increasing order of standard deviations. (Hint: some of the
graphs may have equal standard deviations.)
a<d<b=c
2. The Amazonian Anaconda (snake) grows to an average weight of 330 pounds,
with a standard deviation of 20 pounds. (Hint: think about whether there might be
a quicker way to do the following problems.)
(a) What percent of adult Anacondas weigh between 270 and 390 pounds?
Use Empirical Rule:
Mean ± 3 standard deviation => 99.7%
(b) The largest 16% of Anacondas will be heavier than what weight?
Mean + 1 standard deviation=330+20=350 pounds
(c) The smallest 2.5% of Anacondas will be smaller than what weight?
Mean - 2 standard deviation=330-2*20=290 pounds
 Forward Normal Distribution Problem
3. In 2002, the scores of the students taking the SATs were approximately normal
with mean 1020 and standard deviation 207.
(a) What is the standardized score for an SAT score of 620?
𝟔𝟐𝟎 − 𝟏𝟎𝟐𝟎
= −𝟏. 𝟗
𝟐𝟎𝟕
(b) What does the standardized score mean in terms of the story?
how many standard deviations away from the mean our specific SAT score is
(c) What percent of all students taking the SATs in 2002 had scores less than 620?
Let X be the scores of the students taking the SATs in 2002
P(X < 620) = P(Z < -1.9)=2.87%
(d) What percent of all students taking the SATs in 2002 had scores higher than
1100?
Standard score
𝟏𝟏𝟎𝟎−𝟏𝟎𝟐𝟎
𝟐𝟎𝟕
= 𝟎. 𝟒
=> P(X > 1100)=1-P(X < 1100)=1-P(Z < 0.4)=100%-65.54%=34.46%
(e) Sketch a Normal curve for part (d) on which the mean 1020 and standard
deviation 207 are correctly located. Shade the appropriate area that shows the
percent of students with SAT scores higher than 1100.
Show the mean 1020 in the middle and mark mean ± 1, 2, 3 standard
deviations. Draw a line at 1100 with shading to the right of that line and
labeled with shaded area =34.46%.
(f) Sketch a Normal curve for part (d) in terms of Z. (Sketch Z = 0, the standardized
score, and shade the appropriate area that shows the percent of students with SAT
scores higher than 1100.)
Show the mean Z=0 in the middle and mark mean ± 1, 2, 3 standard
deviations. Draw a line at 0.4 with shading to the right of that line and
labeled with shaded area =34.46%.
4. The Wechsler Adult Intelligence Scale (WAIS) is an IQ test. Scores on the WAIS
for the 20 to 34 age group are approximately Normally distributed with mean 110
and standard deviation 15. Scores for the 60 to 64 age group are approximately
Normally distributed with mean 90 and standard deviation 14. Sarah, who is 30,
scores 130 on WAIS. Her mother, who is 60, takes the test and scores 115.
(a) Express both scores as standardized scores that show where each woman stands
within her own age group.
Sarah:
𝟏𝟑𝟎−𝟏𝟏𝟎
𝟏𝟓
= 𝟏. 𝟑
Her mother:
𝟏𝟏𝟓−𝟗𝟎
𝟏𝟒
= 𝟏. 𝟖
(b) Who scored better compared to their respective age groups?
Sarah’s mother scored higher relative to her age group
(c) Sarah scored higher than what percent of all test takers within her age group? How
about her mother within 60 to 64 age group?
Sarah :
X= Scores on the WAIS for the 20 to 34 age group ~N (𝛍 = 𝟏𝟏𝟎, 𝛔 = 𝟏𝟓)
P(X < 130) = P(Z < 1.3) = 90.32%
Her mother:
Y = Scores on the WAIS for the 60 to 64 age group ~N(𝛍 = 𝟗𝟎, 𝛔 = 𝟏𝟒)
P(Y < 115) = P(Z < 1.8) = 96.41%