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 th

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

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Bơ 3 years 2021-08-27T07:19:17+00:00 1 Answers 65 views 0

Answers ( )

  1. thuthao
    0
    2021-08-27T07:20:46+00:00
    Reply

    Answer:

    39.98

    Step-by-step explanation:

    40.55+41.51+42.01+38.76+34.28+35.38+37.48+40.87+50.32+41.59+42.71=

    479.78

    479.78/12=39.98

    I hope this helps:

     Without the outlier:

    40.55+41.51+42.01+38.76+34.28+35.38+37.48+40.87+41.59+42.71= 395.14

    395.14/11= 39.52

    the answer you wanted:

    39.98-39.52= 0.46

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