ACount the days which had 1 student tardy. Divide this frequency by 30 to get the relative frequency. Do the same for the rest of the days.
BThe mean of the data set is the sum of the values divided by the total number of values in the set. The standard deviation of a sample is calculated similarly to as the standard deviation of an entire set, but we need to divide by n-1 instead of n in the square root.
CUse the mean and the standard deviation you calculated in Part B. Remember to set Xscl to 1 in order to graph both the histogram and the distribution on a single graph properly.
DUse the ShadeNorm function to shade the area under the distribution. Set the upper limit to 5 students.
ESince we cannot set the upper limit to infinity, use an arbitrary big number for the upper limit. You can use 10^(99) for example.
ASee solution.
BMean: 2.73 Standard Deviation: about 1.143
CSee solution.
DAbout 11 %, see solution.
EAbout 13.33 %, see solution.
Practice makes perfect
To graph a relative frequency histogram, it might be useful to find the frequencies first. Let's count the number of ones in the data set.
2
4
3
3
4
2
1
3
3
4
3
2
2
3
1
2
3
3
3
0
2
3
2
5
5
3
3
3
4
1
There are 3 ones in the data set. Let's find the frequencies for other numbers of tardy students!
Number of Tardies
Frequency
0
1
1
3
2
7
3
13
4
4
5
2
To find the relative frequencies, we have to divide the frequencies by the number of all data values. In this case, we have 30 data points. Let's divide!
Number of Tardies
Frequency
Relative Frequency
0
1
≈ 0.033
1
3
0.1
2
7
≈ 0.233
3
13
≈ 0.433
4
4
≈ 0.133
5
2
≈ 0.067
Finally, we can use the first and the second column in our table to help us create the relative frequency histogram on our calculator. We start by pushing the STAT button. Choose Edit and enter your values.
Once the values have been entered, we can plot them by pushing 2nd, Y=, and then choosing one of the plots in the list. Make sure you turn the plot ON before you choose the scatter plot type. Then, match the lists for the x- and y-values accordingly.
By pushing GRAPH, the calculator will plot the data. Note that we will need to change the viewing window so that we can see all of the points.
We need to find the mean μ and the standard deviation for the number of tardy students per day given a sample measured by Ms. Greene. The mean of the data set is the sum of the values divided by the total number of values in the set.
μ=∑_()x_()/n
We are told that the sum total is 82. There are 30 values in this data set. Let's find the sample mean!
μ=82/30 ≈ 2.73
We found that the mean is about 2.73 tardy students per day. Next, we want to find the standard deviation of the given sample. Recall the formula for calculating the standard deviation of the entire data set.
Standard Deviation
sqrt((x_1-μ )^2+(x_2-μ )^2+... +(x_n-μ )^2/n)
The standard deviation of a sample is calculated by dividing by n-1 instead of n in the square root.
Standard Deviation of a Sample
sqrt((x_1-μ )^2+(x_2-μ )^2+... +(x_n-μ )^2/n-1)
Let's use the mean we found to find the differences of the number of students and the mean squared.
x_n
x_n-μ
(x_n-μ)^2
2
2-2.73=- 0.73
(- 0.73)^2 = 0.5329
1
1-2.73=- 1.73
(- 1.73)^2 = 2.9929
2
2-2.73=- 0.73
(- 0.73)^2 = 0.5329
3
3-2.73=0.27
0.27^2 = 0.0729
5
5-2.73=2.27
2.27^2 = 5.1529
4
4-2.73=1.27
1.27^2 = 1.6129
3
3-2.73=0.27
0.27^2 = 0.0729
3
3-2.73=0.27
0.27^2 = 0.0729
0
0-2.73=- 2.73
(- 2.73)^2 = 7.4529
3
3-2.73=0.27
0.27^2 = 0.0729
3
3-2.73=0.27
0.27^2 = 0.0729
3
3-2.73=0.27
0.27^2 = 0.0729
1
1-2.73=- 1.73
(- 1.73)^2 = 2.9929
2
2-2.73=- 0.73
(- 0.73)^2 = 0.5329
3
3-2.73=0.27
0.27^2 = 0.0729
3
3-2.73=0.27
0.27^2 = 0.0729
4
4-2.73=1.27
1.27^2 = 1.6129
2
2-2.73=- 0.73
(- 0.73)^2 = 0.5329
3
3-2.73=0.27
0.27^2 = 0.0729
3
3-2.73=0.27
0.27^2 = 0.0729
4
4-2.73=1.27
1.27^2 = 1.6129
3
3-2.73=0.27
0.27^2 = 0.0729
3
3-2.73=0.27
0.27^2 = 0.0729
2
2-2.73=- 0.73
(- 0.73)^2 = 0.5329
4
4-2.73=1.27
1.27^2 = 1.6129
2
2-2.73=- 0.73
(- 0.73)^2 = 0.5329
2
2-2.73=- 0.73
(- 0.73)^2 = 0.5329
3
3-2.73=0.27
0.27^2 = 0.0729
5
5-2.73=2.27
2.27^2 = 5.1529
1
1-2.73=- 1.73
(- 1.73)^2 = 2.9929
Sum of Values
= 37.867
Finally, we need to divide by 30-1 and then calculate the square root. We should once again round to the nearest tenth.
Standard Deviation: sqrt(37.867/30-1)≈ 1.143
The standard deviation is about 1.143 students.
We want to sketch the normal distribution with a mean of 2.73 and the standard deviation σ of 1.143. Let's once again use a graphing calculator for help. First we push Y= and use the function normalpdf in the first row.
Then, by pushing GRAPH, the calculator will plot the distribution. Note that the bin width Xscl needs to be 1 for us to graph the distribution over the relative frequency histogram.
We want to estimate the percentage of days on which 4 people were tardy. To do it, we can use a graphing calculator. On your calculator push 2nd, then VARS, choose the second option. Enter the distribution's mean, standard deviation, and the lower and upper limits. For our case, we should set the lower limit to 4 and the upper limit to 5.
Having entered all of the values, push ENTER until we get a result.
Therefore, about 11 % of days had 4 students tardy. We can see this portion on the graph by using the ShadeNorm function.
We will follow a similar procedure like the one in Part D. However, this time we will set an arbitrary big number for the upper limit of tardy students. Let's use 10^(99) as an example.
Having entered all of the values, push ENTER until we get a result.
Therefore, about 13.33 % of days had 4 or more students tardy. We can also see this portion on the graph by using the ShadeNorm function.
