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Topics 15 – 20
Solutions
1.
An educational testing corporation has designed a standard test of mechanical aptitude. Scores on this test
are normally distributed with a mean of 75 and a standard deviation of 15.
a.
If a subject is randomly selected and tested, find the probability that his score will be between 75
and 100.
100  75
 1.67
15
75  75
Z2 
0
15
P(75  x  100)  P(0  Z  1.67)  .453
Z1 
b.
If a subject is randomly selected and tested, find the probability that his score will be below 85.
85  75
 0.67
15
P( x  85)  P( Z  0.67)  .748
Z
c.
If a subject is randomly selected and tested, find the probability that his score will be above 80.
80  75
 0.33
15
P( x  80)  P( Z  0.33)  .37
Z
d.
Find the score a subject must score in order to be considered better than 60 percent of the
population (this is also called the 60th percentile)
P( Z *  Z )  0.60
Z *  .25
x  75
.25 
15
x  78.8
e.
If we were to look at samples of 100 people taking this test, what would the Standard deviation of
the sample means be?
 x  15
x 
f.
15
100
 1 .5
What would the probability be of a sample mean being at least 80 if we have a sample of 100
people taking the test?
80  75
 3.33
1.5
P( x  80)  P( Z  3.33)  .00043
Z
g.
We sampled 100 people and found a mean of 72. What is the probability that a sample at most
this average occurs by chance?
72  75
 2.00
1.5
P( x  72)  P( Z  2.00)  .0228
Z
Topics 15 – 20
Solutions
2.
In one county, the conviction rate for speeding is 85%. Suppose we look at 100 summonses. (20 pts)
a. What is the parameter of interest?
The conviction rate for speeding tickets in the given county
b. What is π?
π = .85
c. What is σ for the 100 summonses picked?
 pˆ 
d.
.85(1  .85)
 .0357
100
What is the probability that there will be at least 90 convictions in the sample?
.9  85
 1.4
.0357
P( pˆ  90%)  P( Z  1.4)  ..0807
Z
e.
What is the probability that we will have between 75 and 80 convictions in the sample?
.75  85
 2.80
.0357
.80  85
Z2 
 1.40
.0357
P(75%  pˆ  80%)  P(2.80  Z  1.4)  .078
Z1 
Topics 15 – 20
Solutions
3.
In a randomly selected paragraph of text that had 67 words in it from a Charles Dickens novel, the average
length of each word was found to be 3.87 characters, with a sample standard deviation of 1.95. We are
interested in the value of mean of the length of all words in a Dickens novel. For this problem assume that a
paragraph is a reasonable representation of the entire text (30 pts)
a. Find a 90% confidence interval for the mean of the length of all words of a Dickens novel. Be sure
to show your work and include the following.
 State the variable and parameter of interest and whether the variable is quantitative or
categorical
 Find all the necessary statistical information needed to answer this question
 Show all pertinent technical conditions are satisfied
 Use proper notation
 Include a sketch of your results.
 State the confidence interval in the context of the problem.
b. The true mean for a certain Dickens novel is 4.23, is this in the interval you created?
Take inventory:
n  67
df  67  1  66
x  3.87
S x  1.95 (Since we do not know  we will use a t  interval )
SE x 
1.95
 .238
67
x = the number of characters in a particular word in a Dickens Novel
μ = the average number of characters per word in the entire text
The variable is quantitative
Technical Conditions:
67 ≥ 30 so n is large enough, but it is questionable that we use a single
paragraph to represent the text – though this was explained away in the
problem declaration. Therefore, technical conditions are met.
Finding the 90% Confidence interval for the value of μ:
t * n 1  t 60  1.671 (Using the table)
90% CI : x  t * n 1 ( SE x )
3.87  1.671(.238)
3.87  .398
(3.472,4.268)
We are 90% confident that the true mean number of characters in a word in this particular
Dickens novel is within the interval 3.472 and 4.268 characters.
The value 4.23 is contained in our confidence interval.
Topics 15 – 20
Solutions
4.
The school’s cafeteria wants to know the milk preference of lunch purchasers at FHS. A random sample
of students was taken. When asked their preference, it was found that of the 33 sampled 23 preferred
chocolate milk. (30 pts)
a.
b.
Find the 90% confidence interval of the proportion of students that prefer Chocolate milk. Be sure
to include the following
 State the variable and parameter of interest and whether the variable is quantitative or
categorical
 Find all the necessary statistical information needed to answer this question
 Show all pertinent technical conditions are satisfied
 Use proper notation
 Include a sketch of your results
 State the confidence interval in the context of the problem.
Would we be surprised if the true proportion of Chocolate milk drinkers is 50%, 55% or 60%?
Take Inventory:
x  23
n  33
23
 .697
33
 ?
pˆ 
SE pˆ 
.697(1  .697)
 .080
33
Variable of interest: (x) A students preference for milk in the school cafeteria
Parameter of interest: (π) the true proportion of students that prefer chocolate milk
Create a 90% Confidence interval for π:
pˆ  Z * ( SE pˆ )
.697  1.645(.080)
(.5654,.8286)
We are 90% confident that the true proportion of students in our cafeteria that prefer chocolate milk is in the interval
(.5654, .8286)
Would we be surprised?
π=50%: We would be very surprised since this not close to our interval
π=55%: We would be fairly surprised since it falls just outside our interval
π=60%: We would not be surprised since this value is well within our interval