Question

On the first of each month, stacie runs a 5k race. She keeps track of her times to track her progress. Her time in minutes is recorded in the table: jan 41. 55 july 36. 38 feb 42. 51 aug 38. 48 mar 43. 01 sept 41. 87 apr 39. 76 oct 51. 32 may 37. 32 nov 42. 59 june 35. 28 Dec 43. 71 determine the difference between the mean of the data, including the outlier and excluding the outlier. Round to the hundredths place. 1. 21 0. 93 40. 98 40. 22.

Answer 1

Difference between the mean of the data, including the outlier and excluding the outlier is;

B: 0.93

Mean average

Her time in minutes for the given months are;

• January = 41.55 minutes
• July = 36. 38 minutes
• Feb = 42.51 minutes
• Aug = 38.48 minutes
• Mar = 43. 01 minutes
• Sept = 41.87 minutes
• Apr = 39. 76 minutes
• Oct = 51.32 minutes
• May = 37. 32 minutes
• Nov = 42.59 minutes
• June = 35.28 minutes
• Dec = 43.71 minutes

Looking at all the numbers, they largely range from 36 to 44 which means that the outlier is 51.32

Total number of minutes with outlier = 493.78 minutes

Number of months = 12

Thus;

Mean with outlier = 493.78/12

Mean with outlier = 41.15 minutes

Mean without outlier = (493.78 - 51.32)/(12 - 1)

Mean without outlier = 40.22 minutes

Thus;

Difference in both means = 41.15 - 40.22

Difference = 0.93

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