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Module 2 Exam Quiz Study Guide
Exam Page 1
During an hour at a fast food restaurant, the following types of sandwiches are ordered:
Cheeseburger Fish Cheeseburger Hamburger Fish Chicken Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Chicken Fish Turkey Fish Hamburger Fish Cheeseburger Fish Cheeseburger Hamburger Fish Fish Cheeseburger Hamburger Fish Turkey Turkey Chicken Fish Chicken Cheeseburger Fish Turkey Fish Fish Hamburger Fish Fish Turkey Chicken Hamburger Fish Cheeseburger Chicken Chicken TurkeyFish Hamburger Chicken Fish
a) Make a frequency distribution for this data.
Sandwiches Frequency Fish 20 Chicken 9 Hamburger 9 Cheeseburger 8 Turkey 7 Total 53
b) Make a relative frequency distribution for this data. Include relative percentages on this table.
Copied and pasted from answer above to save on time not having to re-type
Sandwiches Calculation Relative Frequency Relative Percentage
Fish 20/53 = 0.3773 x 100 = 37.73 = 38% Chicken 9/53 = 0.1698 x 100 = 16.98 = 17% Hamburger 9/53 = 0.1698 x 100 = 16.98= 17% Cheeseburger 8/53 = 0.1509 x 100 = 15.09 = 15% Turkey 7/53 = 0.1320 x 100 = 13.2 = 13% Total 53 1 100%
During an hour at a fast food restaurant, the following types of sandwiches are ordered:
Cheeseburger Fish Cheeseburger Hamburger Fish Chicken Hamburger Cheeseburger Fish Hamburger Turkey Fish Chicken Chicken Fish Turkey Fish Hamburger Fish Cheeseburger Fish Cheeseburger Hamburger Fish Fish Cheeseburger HamburgerFish Turkey Turkey Chicken Fish Chicken Cheeseburger Fish Turkey Fish Fish Hamburger Fish Fish Turkey Chicken Hamburger Fish Cheeseburger Chicken Chicken TurkeyFish Hamburger Chicken Fish
a) Make a frequency distribution for this data.
Major Frequency Hamburger 9 Cheeseburger 8 Fish 20 Turkey 7 Chicken 9
Total 53
b) Make a relative frequency distribution for this data. Include relative percentages on this table.
Answer Key
Exam Page 3
Consider the following data: {22, 18, 16, 26, 20, 24} a) Find the sample mean of this data.
22+18+16+26+20+24 = 126 n = 6
xbar = ∑xi / n ∑xi = 126 n = 6
126/6 = 21
sample mean = 21
b) Find the range of this data.
range = highest value - lowest value high value = 26 low value = 16
26-16 = 10
Range = 10
c) Find the sample standard deviation of this data.
s^2 = variance s = standard deviation
s^2 = ∑(xi-xbar)^2 / (n-1)
Therefore, the 60th percentile index for this data set is the 9th observation. In the list above, the 9th observation is 417.
∑(xi-xbar)^2 / (n-1) =
- xi = 16, 18, 20, 22, 24,
- xbar =
- n =
- (16-21)^2 =
- (18-21)^2 =
- (20-21)^2 =
- (22-21)^2 =
- (24-21)^2 =
- (26 -21)^2 =
- 6-1 = n-1 =
- 70/5 =
- s^2 (variance) =
- √14 = 3. standard deviation (s) = √variance
- standard deviation of sample = 3.
- coefficient of variation = (standard deviation / mean) x d) Find the coefficient of variation.
- standard deviation = 3.
- sample mean =
- (3.74/21) x 100 = 17.
Exam Page 4
Suppose that you have a set of data that has a mean of 58 and a standard deviation of 8. a) Is the point 50 above, below, or the same as the mean. How many standard deviations is 50 from the mean.
z = (x-u) / o x = 50 u = o = 8
(50-58) / 8 = - 1
z = - 1 point 50 is 1 standard deviaition below the mean below the mean because negative
b) Is the point 42 above, below, or the same as the mean. How many standard deviations is 42 from the mean.
z = (x-u) / o x = u = 58 o = 8
(42-58) / 8 = - 2
z = - 2 point 42 is 2 standard deviations below the mean below the mean because negative
Suppose that you have a set of data that has a mean of 58 and a standard deviation of 8. a) Is the point 50 above, below, or the same as the mean. How many standard deviations is 50 from the mean.
c) Is the point 54 above, below, or the same as the mean. How many standard deviations is 54 from the mean.
z = (x-u) / o x = 54 u = 58 o = 8
(54-58) / 8 = -0.
z = -0. point 54 is 0.5 (1/2) standard deviations below the mean below the mean because negative
d) Is the point 84 above, below, or the same as the mean. How many standard deviations is 84 from the mean.
z = (x-u) / o x = u = 54 0 = 8
(84-54) / 8 = 3.
z = 3. point 84 is 3.75 standard deviations above the mean above the mean because positive
-2.0 points
Instructor Comments In part c the mean is 58, you are using 54 in your calculation.
Answer Key
Exam Page 5
Consider the following set of data: {20, 5, 12, 29, 18, 21, 10, 15} a) Find the median.
put in order lowest to high
5, 10, 12, 1 5, 18 , 20, 21, 29
because this is an even number data set and there is not an exact middle value, have to add the 2 corresponding middle values and divide by 2 to get median of an even numbered data set
15+18 = 33 33 / 2 = 16.
median = 16.
The z-score is - .5, so the data point 54 is .5 standard deviations below the mean (the negative sign indicates that the point is below the mean).
d) Is the point 84 above, below, or the same as the mean. How many standard deviations is 84 from the mean.
d) The data point 84 is above the mean. Now use the z-score to determine how many standard deviations 84 is above the mean. We are told that the mean is 58 and the standard deviation is 8. So, the z-score is given by:
The z-score is 3.25, so the data point 84 is 3.25 standard deviations above the mean.
Consider the following set of data: {20, 5, 12, 29, 18, 21, 10, 15} a) Find the median.
a) In order to find the median, we must first put the numbers in ascending order:
5, 10, 12, 15, 18, 20, 21, 29.
Notice that there are two “middle” numbers, 15 and 18. The median is the average of these two numbers. Median = (15+18)/2 = 16.5.
b) Find the mode of this set.
b) No number occurs more than once, so there is “no mode”.
b) Find the mode of this set.
no mode. there is not a value that appears more than once in the data set.
Answer Key
