Question

A) (3pts) P(y 2.00) B) (3pts) P(3.33 Sy s 3.75) C) (3pts) The cut off values for the middle 40 percent. D) 3pts) The cut of value for the highest 1 percent. E) (3pts) Calculate the IQR Problem2 number of hours spent in the Santa (I0 pts) Suppose we have the following data available for the average Fe College Math lab. Male Female 1.50 0.75 1.20 0.82 1.30 0.91 1.05 0.84 0.75 0.59 0.80 1.02 1.00 1.90 0.95 2.10 0.925 2.20 2.50 2.50 3.00 3.50 3.50 2.20 4.01 1.80 Use Data Displays to comment on the shape of the distribution, measures of center possible skew, and outliers for both genders. (Hint: Use as many displays as needed. 1) and spread,

Answer 1

> male=c(1.50,1.20,1.30,1.05,0.75,0.80,1.00,0.95,0.925,2.50,3.00,3.50,4.01)

> female=c(0.75,0.82,0.91,0.84,0.59,1.02,1.90,2.10,2.20,2.50,3.50,2.20,1.80)

> library(moments)

## For Male ##

mean(male)

[1] 1.729615

Mean of the male is 1.729615.

> var(male)

[1] 1.261069

## Skewness and kurtosis of male ##

> skewness(male)

[1] 0.9943119

Here the value of skewness B1 and Y1  is less than 1  

i.e. 0.9943. This value implies that the distribution of the data is slightly positively skewed.

> kurtosis(male)

[1] 2.43481

For the kurtosis, we have 2.301051 implying that the distribution of the data is platykurtic, since the computed value is less than 3.

If B2 <3 hence the distribution of the data is platykurtic

hist(male)

From the graph we also seen that the distribution is positively skewed.

> ## For Female ##

mean(female)

[1] 1.625385

Mean of the male is 1.625385

> var(female)

[1] 0.7710603

> skewness(female)

[1] 0.5628414

Here the value of skewness B1 and Y1  is less than 1  

i.e. 0.5628414. This value implies that the distribution of the data is slightly positively skewed.

> kurtosis(female)

[1] 2.451395

For the kurtosis, we have 2.451395 implying that the distribution of the data is platykurtic, since the computed value is less than 3.

If B2 <3 hence the distribution of the data is platykurtic

> hist(female)