A researcher created a dot-plot and a stem-and-leaf plot for the average hours of sleep for 20 high school students who had been randomly selected from a very large high school. They kept a record of the number of hours they slept each night for a week. These seven values were averaged to obtain an average night of sleep for each. The parameters that provide measures of central tendency were estimated. The 20 students slept an average of 9, 8, 8, 7.5, 6, 6, 4, 5.5, 7, 8, 5, 7.5, 6.5, 10, 8.5, 6.5, 5, 5.5, 7, and 7.5 hours. Find mean, median, and mode and state which measure of central tendency is more suitable for the above data. Find the sample range and the sample interquartile range. Find the sample mean absolute deviation. Find the …show more content…
Explain. Solution:- There is an individual series-
As we know
Mean = sum of the observation No of observation
Mean = 9+8+8+7.5+6+6+4+5.5+7+8+5+7.5+6.5+10+8.5+6.5+5+5.5+7+7.5 20 Mean = 138 20 Mean = 6.9 hours …show more content…
4,5,5, 5.5 , 5.5 , 6, 6, 6.5 ,6.5 ,7 ,7 ,7.5 ,7.5 ,7.5 ,8 ,8 ,8, 8.5 , 9, 10
Here no. of observation is even so median is Arithmetic mean of (N/2)th term +(N/2 +1)th term
= (20/2)th term +(20/2 +1)th term
= Arithmetic mean (10th term +11th term)
= 7+7 2
= 14÷2
= 7
Mode
Here 7.5 and 8 both has same frequency
It is bi-model. Modes are 7.5 and 8

Sample range
As we know range is difference between largest term and lowest term
Range = L- S = 10-4 = 6
Sample interquartile range
Sample interquartile range = Q3 -Q1 2
Q3(Third quartile) = 3( N/4)th term (Here n is even no)
Q1 (First quartile) = (N/4)th term
Sample interquartile range = 3×20/4 – 20/4 2 = 15 – 5 2 = 5

Sample mean absolute deviation.

Mean=6.9 (as per a. part)
Mean deviation =