The number of customers, cc, served in a day at a counter for 40 days are given below.

  1. (a)

    Calculate the mean number of customers served.

    Answer ... customers

  2. (b)

    Calculate the standard deviation.

    Answer ... customers

  3. (c)

    If 3 more customers were added to the counter each day, how would the mean and standard deviation be affected by this change?

    Answer

    Mean ...

    Standard deviation ...

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Worked solution

20(a)

1. Find class midpoints and compute the weighted total

Class midpoints: 2.5, 7.5, 12.5, 17.52.5,\ 7.5,\ 12.5,\ 17.5

∑fx=11(2.5)+18(7.5)+9(12.5)+2(17.5)=310\sum fx=11(2.5)+18(7.5)+9(12.5)+2(17.5)=310

2. Divide by the total frequency to find the mean

Total frequency =40=40.

Mean=31040=7.75\text{Mean}=\frac{310}{40}=7.75

Answer

8.758.75 customers

20(b)

1. Use the class midpoints to find $\sum fx^2$

The class midpoints are 2.5, 7.5, 12.5, 17.52.5,\ 7.5,\ 12.5,\ 17.5.

∑fx2=11(2.5)2+18(7.5)2+9(12.5)2+2(17.5)2\sum fx^2=11(2.5)^2+18(7.5)^2+9(12.5)^2+2(17.5)^2

=68.75+1012.5+1406.25+612.5=3100=68.75+1012.5+1406.25+612.5=3100

2. Substitute into the standard deviation formula

Using ∑f=40\sum f=40 and the grouped-data mean xˉ=7.75\bar{x}=7.75,

σ=∑fx2∑f−xˉ2\sigma=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}^2}

=310040−(7.75)2=\sqrt{\frac{3100}{40}-(7.75)^2}

=17.4375=4.1758…=\sqrt{17.4375}=4.1758\ldots

σ≈4.18\sigma\approx 4.18

Answer

4.184.18 customers

20(c)

1. Determine the new mean after adding 3 customers each day

Adding 33 customers to each day's number increases every data value by 33.

So the mean also increases by 33.

8.75+3=11.758.75+3=11.75

2. Determine the new standard deviation after adding 3 customers each day

Adding the same number, 33, to every data value does not change the spread of the data.

So the standard deviation remains unchanged.

4.184.18

Answer

Mean: 11.7511.75 customers

Standard deviation: 4.184.18 customers

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