Question

3. A. The following are a list of 12 prices for Ford stock in 2018: 13.2,...

3. A. The following are a list of 12 prices for Ford stock in 2018: 13.2, 10.6,10.8, 10.8, 11.4, 10, 9.3, 11.7, 11.05, 8.6, 9.4, 7.81. Describe this data using either a histogram or a boxplot. Be sure to show/explain how you obtain the numbers used in your graph.

B. The table below shows the prices of Ford stock on 3 different days. Calculate the mean and the standard deviation of the Ford stock prices.

Month

Ford

Toyota

Jan

13.20

150

July

11.05

130

Dec

7.81

110

  1. The mean Toyota price $130 is and the standard deviation is $20. Calculate the correlation between the Ford and Toyota stock prices during this period.

What proportion of Staten Island residents have done Winter sessions?

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Answer #1

3.

A.

Box plot for the given 12 prices for Ford stock:

Shown vertically:

13.25 13 12.75 12.5 12.25 12 11.75 11.5 11.25 10.75 10.5 10.25 10 9.75 9.5 9.25 8.75 8.5 8.25 7.75 Population size: 12 Median

Shown horizontally:

Boxplot 1 title: Ford prices Min: 7.81 Q1: 9.325 Median: 10.7 Q3: 11.3 125 Max: 13.2 Boxplot limits: Overall min: 7 Overall max: 14 Distance between tick marks: 3 Axis Title: Display Numbers on Boxplot: * Image Size: Width-550 Height-200 Draw here Copy to Clipboard Ford price 7.81 9.325 10.7 I1.3125 13.2 10 13

The data has no outliers and the distribution of the data is negatively skewed because the median, M is closer to the upper quartile, Q3. (For a distribution which is negatively skewed, the box plot will show the median closer to the upper quartile).

B.

Ford prices: 13.20, 11.05, 7.81. So, sample size, n= 3

Mean, ar{X} =Sigma X/n =(13.20+11.05+7.81)/3 =10.69

Standard deviation, s =Σ(X-X)2/(n-1) =2.71

C.

Let X represent Ford prices and Y represent Toyota prices.

Std.deviation of Ford prices, Sx =2.71

Std.deviation of Toyota prices, Sy =20

X Y X-ar{X} Y-ar{Y} (X - X)(Y- Y)​​
13.20 150 2.51 20 50.2
11.05 130 0.36 0 0
7.81 110 -2.88 -20 57.6
ar{X}=10.69 ar{Y}=130

Sigma [(X-ar{X}) (Y-ar{Y})] =

107.8

Covariance of X and Y =Cova(X,Y) =Sigma [(X-ar{X}) (Y-ar{Y})]/(n-1) =107.8/2 =53.9

Now, the correlation between Ford and Toyota prices is: r =Cova(X,Y)/(Sx. Sy) =53.9/[(2.71)(20)] =0.99, that is, almost 1. So, there is a perfect positive correlation between Ford and Toyota prices.

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