The following information concerning grades is posted on the bulletin board.
Test Grade 55 65 75 85 95
Z-Score -2 -1 0 1 2
(a) What is the average test grade?
(b) What test score has a z-score of -2.76?

Tammy and Jamey both applied for the same job. Tammy scored 80 on a
provincial aptitude test where the mean was 70 and the standard deviation
was 4.2. Jamey scored 510 on the company exam where the mean was 490
and the standard deviation was 10.3. Assuming the company uses these test
results as the only criteria for hiring new employees and that both tests are
considered to be equal by company officials, who might get the job?
Explain your answer.

The following are the 41 52 46 42 46 36 44 68 58 44
number of steak 49 48 48 65 52 50 45 72 45 43
dinners served on 50 47 49 57 44 48 49 45 47 48 43
consecutive Sundays 45 56 61 54 51 47 42 53 44 45
at a restaurant. 58 55 43 63 38 42 43 46 49 47
Draw a suitable histogram that has five bars.
Is this data normally distributed?

41 52 46 42 46 36 44 68 58 44
49 48 48 65 52 50 45 72 45 43
47 49 57 44 48 49 45 47 48 43
45 56 61 54 51 47 42 53 44 45
58 55 43 63 38 42 43 46 49 47

The diagram shows a normal distribution with a mean of 28 and a standard
deviation of 4. The values represent the number of standard deviations above
and below the mean. Replace the numbers with raw scores.

The Normal Distribution
The graph at right represents the marks of
a large number of students where the
mean mark is 69.3 percent and the
standard deviation is 7 percent, and the
distribution of marks is approximately
normal.
We know that if the marks are approximately normally distributed, then approximately:
• 68 percent of the marks are between 62.3 percent and 76.3 percent (i.e., µ ± 1σ)
• 34% of the marks are between 69.3 percent and 76.3 percent (i.e., between µ and µ + 1σ)
• 50% of the marks are below 69.3 percent (i.e., less than µ)
• 16% of the marks are above 76.3 percent (i.e., greater than µ + 1σ)

Case 1(a): Calculate the Percentage of Scores Between Two Given Scores
The mean mark for a large number of students is 69.3 percent with a
standard deviation of 7 percent. What percent of the students have a 'B'
mark (i.e., 70 percent to 79 percent)? Assume that the marks are
normally distributed.
Case 1(b): Calculate the Percentage of Passing Scores
The mean mark for a large number of students is 69.3 percent with a standard
deviation of 7 percent. What percent of the students have a passing mark if
they must get 60 percent or better to pass? Assume that the marks are normally
distributed.
Case 1(c): Calculate the Number of Scores
1200 light bulbs were tested for the number of hours of life. The mean life was
640 hours with a standard deviation of 50 hours. Assume that "life in hours" of
light bulbs is normally distributed.
(a) What percent of light bulbs lasted between 600 and 700 hours?
(b) How many light bulbs should be expected to last between 600 and
700 hours?

Case 1(a): Calculate the Percentage of Scores Between Two Given Scores
The mean mark for a large number of students is 69.3 percent with a
standard deviation of 7 percent. What percent of the students have a 'B'
mark (i.e., 70 percent to 79 percent)? Assume that the marks are
normally distributed.