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 50. 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. 93 1. 21.

Answer 1

Outlier is the value in the data set which very different from the other values.the difference between the mean of the data, including the outlier and excluding the outliers 0.93. Thus the option 3 is the correct option

Given information-

Shelly runs a 5 k race on the first of each month.

Mean and Outlier

Mean is the addition of the given data divided by the total number of the values. Outlier is the value in the data set which very different from the other values.

Total values are 12.The addition $s$ of all the values is,

$s=40.55+35.35+41.51+37.48+42.01+40.87+38.76+50.32+36.32+41.59+34.38+42.71$

$s=481.85$

Thus the mean $m$ including the outlier for the given data is,

$m=\dfrac{481.85}{12}$

$m=40.153$

The value 50.32 is much bigger than the other data set. Thus 50.32 is a outlier for the given problem. The addition of all the values excluding the outlier is,

$s_o=40.55+35.35+41.51+37.48+42.01+40.87+38.76+36.32+41.59+34.38+42.71$

$s_o=431.53$

Now the total values are 11. Thus the mean excluding the outlier is,

$m_o=\dfrac{431.53}{11}$

$m_o=39.23$

The difference,

$\Delta m=40.153-39.23$

$\Delta m=0.923$

Thus the difference between the mean of the data, including the outlier and excluding the outliers 0.93. Thus the option 3 is the correct option.

Learn more about the outlier and mean here;

brainly.com/question/1149454