Question

On the first of each month, Shelly runs a 5k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:
Jan 40.55 July 35.38
Feb 41.51 Aug 37.48
Mar 42.01 Sept 40.87
Apr 38.76 Oct 48.32
May 36.32 Nov 41.59
June 34.28 Dec 42.71


Determine the difference between the mean of the data, including the outlier and excluding the outlier. Round to the hundredths place.
39.98
39.22
0.76
1.21

Answer 1

Answer:

0.76

Step-by-step explanation:

The mean of the data including the outlier is:

Mean = (40.55 + 41.51 + 42.01 + 38.76 + 36.32 + 34.28 + 35.38 + 37.48 + 40.87 + 48.32 + 41.59 + 42.71)/12 = 39.98 seconds.

In this case, the outlier comes to be the data: 48.32. If we don't consider that data point, the mean equals:

Mean = (40.55 + 41.51 + 42.01 + 38.76 + 36.32 + 34.28 + 35.38 + 37.48 + 40.87  + 41.59 + 42.71)/11 = 39.22

The difference is:  39.98 - 39.22 = 0.76

Answer 2

Answer:

C . 0.76

Step-by-step explanation:

We are given that on the first of each month , Shelly runs a 5 k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:

Jan  40.55 Feb 41.51

March 42.01 Apr 38.76

May 36.32  June 34.28

July 35.38  Aug 37.48

Sept 40.87 Oct 48.32

Nov 41.59 Dec 42.71

We have to find the difference between the mean of the data , including the outlier and excluding the outlier

Outlier: That observation which is different from other observations.

The outlier in the given observations is 48.32 because is different from other observations.

Mean of the data including the outlier

Mean =$\frac{Sum \;of\;observations}{Total\;number\;of\;observations}$

Mean=$\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+48.32+41.59+42.71}{12}$

Mean=$\frac{479.78}{12}$

Mean=39.982

Mean of the data excluding the outlier

Mean=$\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+41.59+42.71}{11}Mean=[tex]\frac{431.46}{11}$

Mean=39.224

Difference between mean of the data including the outlier and excluding the outlier=39.982-39.224=0.758

Difference between mean of the data including the outlier and excluding the outlier=0.76

Answer Answer:

Step-by-step explanation:

We are given that on the first of each month , Shelly runs a 5 k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table:

Jan  40.55 Feb 41.51

March 42.01 Apr 38.76

May 36.32  June 34.28

July 35.38  Aug 37.48

Sept 40.87 Oct 48.32

Nov 41.59 Dec 42.71

We have to find the difference between the mean of the data , including the outlier and excluding the outlier

Outlier: That observation which is different from other observations.

The outlier in the given observations is 48.32 because is different from other observations.

Mean of the data including the outlier

Mean =$\frac{Sum \;of\;observations}{Total\;number\;of\;observations}$

Mean=$\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+48.32+41.59+42.71}{12}$

Mean=$\frac{479.78}{12}$

Mean=39.982

Mean of the data excluding the outlier

Mean=$\frac{40.55+41.51+42.01+38.76+36.32+34.28+35.38+37.48+40.87+41.59+42.71}{11}Mean=[tex]\frac{431.46}{11}$

Mean=39.224

Difference between mean of the data including the outlier and excluding the outlier=39.982-39.224=0.758

Difference between mean of the data including the outlier and excluding the outlier=0.76

Answer :C . 0.76