Question

On the first of each month, Sasha runs a 5k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table: Jan 39. 55 July 34. 38 Feb 40. 51 Aug 36. 48 Mar 41. 01 Sept 39. 87 Apr 37. 76 Oct 50. 32 May 35. 32 Nov 40. 59 June 33. 28 Dec 41. 71 Determine the difference between the mean of the data, including the outlier and excluding the outlier. Round to the hundredths place. 38. 98 38. 22 1. 21 1. 1.

Answer 1

The difference of mean for the given set of terms including the outlier and excluding the outlier is 1.01.

Mean

The arithmetic mean for a given set of numbers is defined as the central value for the given set of numbers.

Given that Sasha runs a 5k race. Her time in minutes is recorded in the table:

• Jan 39. 55
• Feb 40. 51
• Mar 41. 01
• Apr 37. 76
• May 35. 32
• June 33. 28
• July 34. 38  
• Aug 36. 48  
• Sept 39. 87  
• Oct 50. 32  
• Nov 40. 59
• Dec 41. 71.

The mean for the given set of terms can be calculated by the total sum of all the terms divided by the total number of terms.

Mean = $\dfrac {39. 55+40. 51+41. 01+37. 76+35. 32+33. 28+34. 38+36. 48+39. 87+50. 32 +40. 59+41. 71} {12}$

Mean = 39.23

Thus the mean of the given set of terms is 39.23.

The outlier term in the given set of terms is 50.32. If we exclude this outlier term, then the mean will be given as below.

Mean = $\dfrac {39. 55+40. 51+41. 01+37. 76+35. 32+33. 28+34. 38+36. 48+39. 87+40. 59+41. 71} {11}$

Mean = 38.22

The difference between the mean including the outlier and excluding the outlier is given below.

Difference = 39.23 - 39.22

Difference = 1.01

Hence we can conclude that the difference of mean for the given set of terms including the outlier and excluding the outlier is 1.01.

To know more about the mean, follow the link given below.

brainly.com/question/12513463.