Question

1. Jennifer scored 90.5 on a particular physics exam. The class mean on the exam was
70.47 with a standard deviation of 7.41. Find the z-score for her result.

1a. On the same exam above, Pete scored 62.0 . Calculate his z-score on the
exam.

1b. Which of these scores would be considered as more unusual (or more extreme)?
Explain.

2. In a distribution that is skewed to the right, what is the relationship of the mean, median,
and the mode? (Circle One.)

a. median > mean > mode

b. mode > median > mean

c. mean > median > mode

d. mode > mean > median

3. The weight (in pounds) is given for
30 newborn babies. Draw a boxplot that summarizes
the data in the box below. Be sure to label things
properly. (You can use your calculator to get the 5-
number summary to get the results you need for the boxplot.)

5.5, 5.7, 5.8, 5.9, 6.1, 6.1, 6.3, 6.4, 6.5, 6.6

6.7, 6.7, 6.7, 6.9, 7.0, 7.0 , 7.0, 7.1, 7.2, 7.2

7.4, 7.5, 7.7, 7.7, 7.8, 8.0, 8.1, 8.1, 8.3, 8.7

4. A severe drought affected several western states for 3 years. A Christmas tree farmer
is worried about the drought’s effect on the size of his trees. To decide whether the growth of the trees
has been retarded, the farmer decides to take a sample of the heights of 25 trees. Typically trees of
this age have a mean height of 65 inches , with a standard deviation of 9 inches . Assuming the
distribution is bell-shaped, where do you expect the middle 95% of the tree heights to fall?

Between-____________ and______________ inches.

5. A company had 80 employees whose salaries are summarized in the frequency
distribution below. Find the mean salary and the standard deviation of the salaries. Consider these 80
employees as a population. (Use the shortcut formula for  (that doesn’t use  ).) Show the full table
you develop for these calculations and then the actual calculations.

Salary (dollar)	Employee
5,001-10,000	19
10,001-15,000	14
15,001-20,000	12
20,001-25,000	16
25,001-30,000	19

Mean________________ Standard Deviation_________________

6. To get the best deal on a microwave oven, Jeremy called six appliance stores and
asked the cost of a specific model he was interested in. The prices he was quoted are given below.
By hand, find the sample mean, standard deviation, and variance. Use the computational formula for
standard deviation. (You can use the 1-variable stats from your calculator to get the sums you need
for these calculations.)

$663, $273, $410, $622, $174, $374

Mean :
Standard. Deviation. :
Variance :

7. The monthly telephone usage (in minutes) of 33 adults is listed below. Find the
interquartile range for the telephone usage of the 33 adults. (Again, you can use the 1-variable
statistics from your calculator to help you out, but show those results.)

154, 156, 165, 165, 170, 171, 172, 180, 184, 185, 187

189, 189, 190, 192, 195, 198, 198, 200, 200, 200, 202

205, 205, 211, 215, 220, 220, 225, 238, 255, 265, 401

IQR :_______

7a. Are there any outliers here? If so, which values? Show calculations to justify
your answers here.




Outlier:____________

Answer 1

Q1.

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 70.4
standard Deviation ( sd )= 7.41
------------------------------------------------------------------------------
a.
Jennifer scored 90.5 on a particular physics exam X = 90.5
P(X = 90.5) = (90.5-70.4)/7.41
= 20.1/7.41= 2.7126
Z = 2.7126
------------------------------------------------------------------------------
Pete scored 62.0, X = 62
P(X = 62) = (62-70.4)/7.41
= -8.4/7.41= -1.1336
Z =-1.1336
------------------------------------------------------------------------------
b.
For Jennifer score,
P(X > 90.5) = (90.5-70.4)/7.41
= 20.1/7.41 = 2.7126
= P ( Z >2.7126) From Standard Normal Table
= 0.0033
For Pete,
P(X < 62) = (62-70.4)/7.41
= -8.4/7.41= -1.1336
= P ( Z <-1.1336) From Standard Normal Table
= 0.1285
we see only 0.33% scores are above Jennifer & it is considered to be as an unusual event as the percentage find is less than 5%