The following data represent the weights (in grams) of a simple random sample of a candy. 0.92 0.90 0.90 0.93 0.92 0.86 0.92 0.86 0.80 0.90 Determine the shape of the distribution of weights of the candies by drawing a frequency histogram and computing the mean and the median. Which measure of central tendency best describes the weight of the candy?

We can use the following R code in order to create the histogram:

> x<-c(0.92, 0.90, 0.90, 0.93, 0.92, 0.86, 0.92, 0.86, 0.80, 0.90)

> hist(x)

And we got the figure attached as the result

And as we can see the shape is not bell shaped so then we can conclude that the distribution is not normal

And replacing we got:

We have 10 observations we can calculate the median with the positions 5 and 6 from the dataset ordered and we got:

And as we can see the median is different from the mean so then the distribution for the data is not normal for this case.

Step-by-step explanation:

For this case we have the following data given:

0.92 0.90 0.90 0.93 0.92 0.86 0.92 0.86 0.80 0.90

We can sort the data on increasing way and we got:

0.80 0.86 0.86 0.90 0.90 0.90 0.92 0.92 0.92 0.93

We can use the following R code in order to create the histogram:

> x<-c(0.92, 0.90, 0.90, 0.93, 0.92, 0.86, 0.92, 0.86, 0.80, 0.90)

> hist(x)

And we got the figure attached as the result

And as we can see the shape is not bell shaped so then we can conclude that the distribution is not normal

We can calculate the mean with the following formula:

And replacing we got:

We have 10 observations we can calculate the median with the positions 5 and 6 from the dataset ordered and we got:

And as we can see the median is different from the mean so then the distribution for the data is not normal for this case.

it would rotate about 180 degrees.