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pishuonlain [190]
3 years ago
9

Your friend, Susannah, manages a sandwich shop. She collects data on the number of turkey sandwiches sold per day for a week. Ex

plain to her how to find the mean, median, mode, and range of the number of turkey sandwiches sold. How would that be helpful when she orders new turkey sandwich makings? Which measure do you think would be the most helpful for her and why? Number of sandwiches sold per day
Monday: 73
Tuesday: 88
Wednesday: 85
Thursday: 88
Friday: 99
Saturday: 134
Sunday: 129

Your response must have MORE than 5 complete sentences for this assignment!​
Mathematics
1 answer:
ExtremeBDS [4]3 years ago
6 0

Susannah's data are illustration of measure of central tendencies

<u>(a) The mean</u>

To calculate the mean, she needs to divide the total number of turkey sandwiches sold, by the number of days.

The total is:

\mathbf{Total = 73 +88+85+88+99+134+129}

\mathbf{Total = 696}

The number of days is:

\mathbf{Day = 7}

So, the mean is:

\mathbf{Mean =\frac{Total}{Days}}

\mathbf{Mean =\frac{696}{7}}

\mathbf{Mean =99.4}

Hence, the mean number of turkey sandwiches sold is 99.4

<u>(b) The median</u>

To calculate the median, she needs to arrange the data in ascending or descending order, then select the middle item.

In ascending order, we have: 73, 85, 88, 88, 99, 129, 134

The middle item is 88

So:

\mathbf{Median = 88}

Hence, the median number of turkey sandwiches sold is 88

<u>(c) The mode</u>

This is the data that occurs most.

From the dataset, 88 appears twice (more than others).

So:

\mathbf{Mode = 88}

Hence, the mode number of turkey sandwiches sold is 88

<u>(d) The range</u>

This is the difference between the highest and least data

From the dataset,

The highest is 134, and the least is 73

So:

\mathbf{Range =134 - 73}

\mathbf{Range =61}

Hence, the range of turkey sandwiches sold is 61

The measure that would be most helpful is the median, because it is not affected by outliers.

Read more about measures of central tendencies at:

brainly.com/question/5495004

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