The waiting time, t in minutes, for 40 customers at two phone service shops, Shop A and Shop B, are as follows. Calculate the mean waiting time for Shop A. Ans: ___ min Explain why your answer to (a)(i) is only an estimate. Answer ... Calculate the standard deviation for Shop A. Ans: ___ min Which shop would you choose to go to? Why? Answer ... A customer went to Shop B and his waiting time is 13.2 min. If his waiting time is added to the data, will the mean and standard deviation of the new set of data increase, decrease or remain the same? Answer The mean will ... The standard deviation will ... A teacher recorded the individual time taken by 200 students to complete a cross-country race. The results are presented in the form of a cumulative frequency curve. the median, Ans: ___ min the 90th percentile, Ans: ___ min the qualifying time to run in the second round if only 30 students qualify for the second round, Ans: ___ min the number of students who took more than 50 minutes but no more than 66 minutes. Ans: ___

1. Find the midpoints of each time interval and calculate the estimated mean.

Midpoints: $2.5,\ 7.5,\ 12.5,\ 17.5$

$\text{Mean}=\frac{(2.5)(5)+(7.5)(12)+(12.5)(8)+(17.5)(15)}{40}$

$=\frac{12.5+90+100+262.5}{40}$

$=\frac{465}{40}=11.625$

1. State why the calculated mean from grouped data is not exact.

The data for Shop A is given in grouped intervals, so the individual waiting times are unknown. In (a)(i), the class midpoints were used to represent all values in each interval, so the mean is only an estimate.

1. Use grouped-data midpoints to calculate the estimated standard deviation.

Midpoints: $2.5,\ 7.5,\ 12.5,\ 17.5$

Using $\bar{x}=11.625$ from part (a)(i),

$\sum fx^2=5(2.5)^2+12(7.5)^2+8(12.5)^2+15(17.5)^2$

$=31.25+675+1250+4593.75=6550$

$\text{Standard deviation}=\sqrt{\frac{\sum fx^2}{\sum f}-\bar{x}^2}$

$=\sqrt{\frac{6550}{40}-(11.625)^2}$

$=\sqrt{163.75-135.140625}$

$=5.348\ldots$

1. Compare the two shops using mean and standard deviation, then justify a choice.

Shop A: mean $=11.625$ min, standard deviation $=5.35$ min.

Shop B: mean $=13.2$ min, standard deviation $=2.17$ min.

Since Shop A has the lower mean waiting time, choose Shop A if the aim is to wait for a shorter time on average.

1. Determine the effect on the mean when adding a value equal to the current mean.

For Shop B, the current mean is $13.2$ min.

The new waiting time added is also $13.2$ min.

Since the added value is equal to the current mean, the mean will remain the same.

2. Determine the effect on the standard deviation when adding a value equal to the mean.

The added waiting time $13.2$ min has deviation

$13.2-13.2=0$

from the mean.

It adds no extra spread, but increases the number of data values. Therefore, the standard deviation will decrease.