11.7 Working with the large data set — Practise with help

Work each question closed book first. Clean the data before you calculate, and say whether the data set is the population or a sample.

Warm-up: remember

Work out $25\%$ of $400$ .
Find the mean of $12, 16, 20, 24$ .
Convert $90$ km/h to km/min.

Answers

$0.25 \times 400 = 100$ .
$\dfrac{12 + 16 + 20 + 24}{4} = \dfrac{72}{4} = 18$ .
Divide by 60: $90 \div 60 = 1.5$ km/min.

If question 3 was hard, revise converting units before you start.

Practise with help

1. A large data set holds 5000 daily records. A simple random sample of 250 records contains 40 with fog. Estimate the number of foggy days in the whole data set.

Hint

Find the sample proportion first, then multiply by the population size.

Next step

The proportion is $\dfrac{40}{250} = 0.16$ . Multiply by 5000.

Full working

\frac{40}{250} \times 5000 = 0.16 \times 5000 = 800.

About 800 days in the data set have fog. The data set is the population, and the 250 records are a sample from it.

2. A data set has 3600 records from one region and 2400 from another. A stratified sample of 150 is taken. Find the number of records from each region.

Hint

The total is $3600 + 2400 = 6000$ . Use proportional allocation for each region.

Next step

For the first region, work out $\dfrac{3600}{6000} \times 150$ . Then do the same for 2400.

Full working

n_1 = \frac{3600}{6000} \times 150 = 90, \qquad n_2 = \frac{2400}{6000} \times 150 = 60.

The two sample sizes add to 150, which checks that the allocation is complete.

3. A rainfall column contains the values $12, 15, 14$ , a blank, and $13$ mm. Explain how to clean the data, and find the mean of the known values.

Hint

The blank is a missing value. Decide whether it can be recovered.

Next step

If the true value cannot be found, leave it out. Find the mean of the four known values.

Full working

The blank is a missing value. If it cannot be recovered, leave it out of the calculation and report that one value is missing. The known values sum to $12 + 15 + 14 + 13 = 54$ over 4 values, so the mean is

\frac{54}{4} = 13.5 \text{ mm}.

4. A histogram of wind speed has frequency density 5 in the class $20$ – $30$ km/h. Find the number of days in the class, and the number of days between 20 and 24 km/h.

Hint

Find the whole class first, then the part from 20 to 24.

Next step

The class width is 10, so the frequency is $5 \times 10 = 50$ . The part from 20 to 24 has width 4.

Full working

\text{frequency} = 5 \times 10 = 50, \qquad \text{part} = 5 \times 4 = 20.

There are 50 days in the class, and 20 of them have wind speeds between 20 and 24 km/h.

5. A sample of 100 days has a mean temperature of $15.2$ degrees Celsius and a standard deviation of $3.4$ degrees. A second sample of 100 days has a mean of $15.2$ degrees and a standard deviation of $7.1$ degrees. Compare the two samples.

Hint

The means are equal, so compare the spreads.

Next step

Decide which sample has temperatures closer to the mean.

Full working

Both samples have the same mean temperature, so their typical values agree. The first sample has the smaller standard deviation, so its temperatures are more consistent and cluster closer to $15.2$ degrees. The second sample varies more, with more very warm and very cool days.

6. A data set records sea temperature in degrees Celsius, but 20 of the 500 values are in degrees Fahrenheit. Explain how to clean the data, and say what you must report.

Hint

Choose one unit and convert the other values to it.

Next step

Use ${}^\circ\mathrm{C} = \dfrac{5}{9}({}^\circ\mathrm{F} - 32)$ to convert the 20 values.

Full working

Convert the 20 Fahrenheit values to Celsius using ${}^\circ\mathrm{C} = \dfrac{5}{9}({}^\circ\mathrm{F} - 32)$ . Then every value shares one unit and the summary statistics can be calculated. Report that 20 values were converted, so a reader can see what changed.