Mary got a sales job at her father's company. She was paid $9 per hour plus a 5%commission on her sales. Last week sheworked 40
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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 =
Mean=
Mean=
Mean=39.982
Mean of the data excluding the outlier
Mean=
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 =
Mean=
Mean=
Mean=39.982
Mean of the data excluding the outlier
Mean=
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