Problem 1
n = 50 students
A = 68 is the average height
A = S/n where S is the sum of the heights
nA = S ... multiply both sides by n
S = nA
S = 50*68 .... plug in given values
S = 3400
The sum of the heights of the fifty students is 3400 inches.
Subtract off 82 and 86, as these heights are no longer considered
S-82-86 = S-168 = 3400-168 = 3232
Now divide that over the new sample size of 48. This is a reduction of 2 from n = 50 earlier. In other words, n-2 = 50-2 = 48.
B = new average
B = (S - 82 - 86)/(n - 2)
B = (S - 168)/(n - 2)
B = 3232/48
B = 67.3333333333333
B = 67.33 <<--- round to nearest hundredth
--------------
note: 67 inches is 5 ft, 7 inches
===============================================
Problem 2
To get the mean, add up the values and divide by 13 (since there are n = 13 items in the list)
(27+16+8+5+19+14+22+32+13+5+23+16+8)/13 = 208/13 = 16
The mean = 16.
Now onto the median. First we need to sort the given data set
{27,16,8,5,19,14,22,32,13,5,23,16,8}
to get
{5,5,8,8,13,14,16,16,19,22,23,27,32}
There are 13 items in the list. The middle value will have 6 items to the left of it and 6 items to the right of it. The middle value is in slot 7, so the median is the value 16 (the first copy of 16)
After getting the median, we can split the list into two smaller lists
L = {5,5,8,8,13,14}
U = {16,19,22,23,27,32}
The list L is the list of values lower than the median. The list U is the upper list of values larger than the median. We do not include the median itself in either list L or list U.
The median of the set {5,5,8,8,13,14} is 8. It is the halfway point between the tied middle values of 8 and 8, so (8+8)/2 = 8. This is the value of Q1, so Q1 = 8.
The median of set U = {16,19,22,23,27,32} is 22.5, which is found by averaging 22 and 23 as those are the tied values in the middle. Therefore Q3 = 22.5
The interquartile range (IQR) is
IQR = Q3 - Q1
IQR = 22.5 - 8
IQR = 14.5
--------------
===============================================
Problem 3
This distribution is bimodal because there are 2 modes. The two modes are the values that repeat the most times, and they are tied for that. Specifically the values 1 and 3 are the mode. They show up 4 times each. We can say the mode is {1,3}.
Also, the distribution is a bit skewed to the right. The slight outliers on the right side pull on the right tail to stretch it out and skew it in that direction.
===============================================
Problem 4
Question: How many drinks were sold in the three-hour time period?
Explanation: Count all of the dots in the dot plot and you should find there are 20 dots total. You can also think of each column separately and add up the values in each column 4+2+4+2+3+2+1+1+1 = 20
--------
Question: How many drinks were sold at the 5th price?
Explanation: There are 3 dots over the label "5" on the horizontal axis.
--------
Question: Assuming that the drink prices were plotted from smallest to largest as the horizontal axis increases, were more drinks sold at a higher price or lower price?
Explanation: Much of the points are concentrated toward the left side of the horizontal axis, meaning that more people bought the cheaper drinks (since price #1 is cheaper than price #2, and #2 is cheaper than #3, etc etc). This is another way to see that the data is skewed to the right. The outlier(s) determine the skew direction.