Question

The number of hours Layla worked each week for the first eight weeks of summer break are recorded below.

12, 22, 19, 28, 16, 14, 32, 21

What would happen to the data distribution if Layla works 20, 31, 24, and 25 hours for the last four weeks of summer break?


A. The value of the center decreases and the distribution has a larger spread.

B. The value of the center decreases and the distribution has a smaller spread.

C. The value of the center increases and the distribution has a larger spread.

D. The value of the center increases and the distribution has a smaller spread.

Answer 1

D. The value of the center increases and the distribution has a smaller spread.

Explanation:

The given group of data is

Group 1: 12, 22, 19, 28, 16, 14, 32, 21.

Then, with the change

Group 2: 12, 22, 19, 28, 20, 31, 24, 25.

Let's compare both groups. The range of the first group is

`R_(1)=32-12=20`

The range of the second group is

`R_(2)= 31-12=19`

As you can observe the spread is smaller by 1 unit, because the range indicates the separation or the spread of a distribution. So, the possible correct choices are B and D.

Now, let's find the median to each group. Remember that the median is the value placed right in the middle. In this case, we need to use the mean to find it.

First, we need to order each group.

Group 1: 12, 14, 16, 19, 21, 22, 29, 32.

Group 2: 12, 19, 20, 22, 24, 25, 29, 31.

`M_(1)=(19+21)/(2)=(40)/(2)=20\\ M_(2)=(22+24)/(2)=(46)/(2)=23`

As you can observe, the median is higher for the second group, that means the central value increases.

Therefore, the right answer is D. The value of the center increases and the distribution has a smaller spread.

Answer 2

Answer: The value of the center increases and the distribution has a smaller spread.

Explanation:

Re - arranging the first value in ascending order , we have :

12 ,14, 16 , 19, 21 , 22 , 28 , 32

Median = 19 + 21 / 2

= 40/2

= 20

Also calculating the standard variation in order to know the spread , we need to first of all calculate the mean

Mean = 12 + 14 + 16 + 19 + 21 + 22 + 28 + 32 / 8

Mean = 164/8

Mean = 20.5

Therefore to calculate the standard deviation, we need to calculate the variance, which is

`(12-20.5)^(2)` +
`(14-20.5)^(2)` +
`(16-20.5)^(2)` +
`(19-20.5)^(2)` +
`(21-20.5)^(2)` +
`(22-20.5)^(2)` +
`(28-20.5)^(2)` +
`(32-20.5)^(2)` / 8

Variance = 328/8

Variance = 41

Standard deviation =
`√(variance)`

S.D =
`√(41)`

S.D = 6.4

Also re arranging the second values , we have :

20 , 24 , 25 , 31

Median = 24 + 25 / 2

Median = 49/2

Median = 24.5

Calculating the S.D using the same format , the S.D = 3.9

Comparing the two we can conclude that the value at the center increases and the distribution has a smaller spread