11.7 Working with the large data set — Practise

Work through these on your own. Clean the data before you calculate, and state the unit in your answer.

The questions and the answers below come from a set that is checked by computer every time the book is built. The tables in the questions are small extracts that stand in for your board's data set.

A large data set holds 4000 days of weather records. A simple random sample of 200 days contains 50 wet days. Estimate the number of wet days in the whole data set.
A data set contains 1200 days from station A and 800 days from station B. A stratified sample of 100 days is taken. Find the number of days from station A.
An extract groups daily rainfall into $0$ – $2$ mm with frequency 30 and $2$ – $10$ mm with frequency 40. Find the frequency density of the $2$ – $10$ mm class.
A histogram of daily sunshine has frequency density 6 in the class $10$ – $20$ hours. Find the frequency in the class.
A histogram of daily rainfall has frequency density 3 in the class $0$ – $10$ mm. Find the number of days with rainfall between $4$ mm and $10$ mm.
Six daily maximum temperatures, in degrees Celsius, are $18, 20, 22, 19, 21, 20$ . Find the mean.
Find the standard deviation of $18, 20, 22, 19, 21, 20$ , dividing by $n$ .
Find the interquartile range of $18, 20, 22, 19, 21, 20$ , using the $(n + 1)$ positions.
A data set has $Q_1 = 18.75$ , $Q_3 = 21.25$ and IQR $= 2.5$ . Find the two fences for a possible outlier.
One day in an extract has a missing rainfall value. The five known values are $12, 15, 14, 16, 13$ mm. Find the mean of the known values.
A temperature is recorded as $68$ degrees Fahrenheit, but the rest of the data are in degrees Celsius. Convert $68$ degrees Fahrenheit to degrees Celsius.
A scatter diagram from a large data set plots hours of sunshine against temperature, and a line of best fit passes through $(2, 12)$ and $(6, 20)$ . Find the gradient of the line.

Answers and concise solutions

1. The estimated number of wet days

Answer: The sample proportion is $\dfrac{50}{200} = 0.25$ , so the estimate is $0.25 \times 4000 = 1000$ wet days. Here the data set is the population and the 200 days are a sample from it.

2. The sample size from station A

Answer: The total is 2000 days, so station A contributes $\dfrac{1200}{2000} \times 100 = 60$ days. Stratifying keeps the two stations in proportion.

3. The frequency density

Answer: The class width is $10 - 2 = 8$ , so the frequency density is $\dfrac{40}{8} = 5$ . The two classes have different widths, so a histogram is needed.

4. The frequency

Answer: The frequency is the area of the bar: $6 \times 10 = 60$ days.

5. The number of days

Answer: The part from $4$ mm to $10$ mm has width 6, so the frequency is $3 \times 6 = 18$ days.

6. The mean

Answer: The values sum to 120, and there are 6 of them, so the mean is $\dfrac{120}{6} = 20$ degrees Celsius.

7. The standard deviation

Answer: The mean is 20. The squared deviations sum to 10, so the variance is $\dfrac{10}{6}$ and $\sigma = \dfrac{\sqrt{15}}{3} = 1.29$ (3 s.f.). This is the population convention, dividing by $n$ ; some software divides by $n - 1$ .

8. The interquartile range

Answer: In order the values are $18, 19, 20, 20, 21, 22$ . The position of $Q_1$ is $\dfrac{7}{4} = 1.75$ , giving 18.75. The position of $Q_3$ is $\dfrac{21}{4} = 5.25$ , giving 21.25. The IQR is $21.25 - 18.75 = 2.5 = \dfrac{5}{2}$ .

9. The two fences

Answer: The fences are $18.75 - 3.75 = 15$ and $21.25 + 3.75 = 25$ . A value below 15 or above 25 is a possible outlier in this data set.

10. The mean of the known values

Answer: The known values sum to 70 over 5 days, so the mean is $\dfrac{70}{5} = 14$ mm. Report that one value was missing, so the sample used fewer days.

11. The temperature in degrees Celsius

Answer: Use ${}^\circ\mathrm{C} = \dfrac{5}{9}({}^\circ\mathrm{F} - 32)$ : $\dfrac{5}{9}(68 - 32) = 20$ degrees Celsius. Convert every value to one unit before calculating.

12. The gradient

Answer: The gradient is $\dfrac{20 - 12}{6 - 2} = 2$ , so each extra hour of sunshine is associated with about 2 degrees Celsius more. The correlation is positive; it does not by itself prove that sunshine causes the rise.