Question

1. Women (ages 18 to 25) heights

Data Collection

                                                                                                                    Height (ft.)

1	5.0
2	5.4
3	5.4
4	5.5
5	4.11
6	5.8
7	5.1
8	5.3
9	5.4
10	5.5
11	5.7
12	5.8
13	5.10
14	5.7
15	6.3

13) Plot the third bell curve below:

14) What is the data value associated with a z-score of 0.5? _________________________

15) What is the data value associated with a z-score of -2.4? _________________________

16) What is the variance of this data set? _________________________________________

17) What is the range of data values so that 99% of the data would fall within the mean? ______________________________________________________________________

18) Are there any outliers with this data set? (Outliers refer to data points that lie beyond 3 standard deviation from the mean) _________________________________________________

19) Why does not 100% of the data always fall within 3 standard deviations of the mean? _____________________________________________________________________

Answer 1

Here we have given the data which represent the height of the women in the age group (18 t0 25 )

In order to solve the question we have to first find the mean and the standard deviation of the given data set

So

Mean = Sum of all the observations / Total no.of observations

Mean = 81.11 / 15 = 5.41

As we have given the sample data so we will find the sample variance.

To find this we have to make a table as

Sr.No.
1	5	0.17
2	5.4	0.00
3	5.4	0.00
4	5.5	0.01
5	4.11	1.68
6	5.8	0.15
7	5.1	0.09
8	5.3	0.01
9	5.4	0.00
10	5.5	0.01
11	5.7	0.09
12	5.8	0.15
13	5.1	0.09
14	5.7	0.09
15	6.3	0.80
Total	81.11	3.34

Sample -variance = 3.34 / (15-1) = 3.34 / 14 = 0.24

Sample -variance = 0.24

The sample standard deviation =  

Therefore Sample - sd = 0.49

Now we find the value of x corresponding to the z-score 0.5

The formula to find this is given by

x = 0.5 * 0.49 + 5.41

x = 5.66

Similarly for z-score = -2.4

x = - 2.4 * 0.49 + 5.41

x= 4.23

In the next part, we have asked to find the 99% confidence interval for the population mean

The formula for this is as

(Mean - Z_alpha * sd / n ,  Mean + Z_alpha * sd / n )

Where Z_alpha = 2.58

Hence the 99% confidence interval is

(5.41 - 2.58 * 0.49 / 15 , 5.41 + 2.58 * 0.49 / 15 )

(5.41 -0.08 , 5.41 + 0.08 )

(5.33 , 5.49 )

This is the range (5.33 , 5.49 ) of data values so that 99% of the data would fall within the mean.

In the next part, we have to check that Is any outlier in the data set

For that, we prefer the 3 standard deviation limit. So the data value fall out of the range of 3 standard deviations is treated as an outlier.

(Mean - 3 * standard deviation ,  Mean + 3 * standard deviation)

( 5.41 - 3 * 0.49 , 5.41 + 3 * 0.49 )

(3.94 , 6.88 )

So we observe that there is not any data value falls outside this range. This means there is no outlier in the given data set.

In the next part, we have asked that Why does not 100% of the data always fall within 3 standard deviations of the mean?

Because according to the Chebyshev's theorem there are 89% of the data falls within the 3 standard deviations and according to the normal distribution, there is 99.73% of data will fall within the 3 standard deviations from the mean. This means that there is a chance for the data value to fall outside the 3 standard deviation limits.