Data, sampling and averages Problem set B Worked solutions

A hospital has 360 staff: 150 nurses, 120 doctors and 90 administrators. A survey of 36 staff is needed, with the same proportion of each group. Choosing randomly within each group.

1. (a) Name this sampling method.

   • The population is split into groups and a random sample is taken from each in proportion to its size.
   • This is stratified sampling.

2. (b) How many doctors are in the sample?

   • \(\frac{36}{360}\times 120=12\)

Name the sampling method in each case.

1. (a) All 350 raffle tickets are numbered and a random number generator picks 25 of them.

   • Every member of the population has an equal chance of being chosen, with no pattern: simple random.

2. (b) At a car park, every 8th car that arrives is asked about the parking charge.

   • Items are taken at a fixed interval from a list or a stream: systematic.

3. (c) A market researcher must interview 15 people under 30 and 15 people aged 30 or over. She asks whoever passes until both quotas are full.

   • The sample is filled to fixed numbers in each group, and the interviewer chooses who: quota.

The lengths of long jumps by pupils on a sports day are recorded to the nearest centimetre. The table shows the results.

1. (a) How many pupils were there in total?

   • \(2 + 5 + 9 + 13 + 11 + 6 + 3 + 2 + 1 = 52\)

2. (b) Complete a grouped table for the classes 449.5 ≤ l < 452.5, 452.5 ≤ l < 455.5 and 455.5 ≤ l < 458.5.

   • 449.5 ≤ l < 452.5: 2 + 5 + 9 = 16
   • 452.5 ≤ l < 455.5: 13 + 11 + 6 = 30
   • 455.5 ≤ l < 458.5: 3 + 2 + 1 = 6
   • Check: the total is still 52

3. (c) A recorded value is 455 centimetres. Between which two values does the actual value lie?

   • The data is rounded to the nearest centimetre, so the actual value is within half a centimetre of 455.
   • 454.5 ≤ l < 455.5

Find the mean, the median and the mode of each list.

1. (a) The number of emails received on 7 days: 21, 14, 17, 14, 9, 25, 14.

   • Mean: 114 ÷ 7 = 16.3
   • Ordered: 9, 14, 14, 14, 17, 21, 25
   • Median = 14, mode = 14

2. (b) The shoe sizes of 10 students: 3, 8, 5, 6, 8, 11, 4, 8, 7, 10.

   • Mean: 70 ÷ 10 = 7
   • Ordered: 3, 4, 5, 6, 7, 8, 8, 8, 10, 11
   • Median = 7.5 (halfway between the middle two), mode = 8

Find the value of x.

1. (a) The mean of 5, 8, x, 3x, 14 is 11.

   • \(\frac{5+8+x+3 x+14}{5}=11\)
   • \(27+4x=55\)
   • \(4x=28,\ x=7\)

2. (b) The mean of x + 1, 2x − 3, 3x + 2, 8 is 11.

   • \(\frac{x + 1+2 x - 3+3 x + 2+8}{4}=11\)
   • \(8+6x=44\)
   • \(6x=36,\ x=6\)

The mean mass of 11 eggs is 58.5 g. A twelfth egg is added, and the mean mass of the 12 eggs is 59.2 g.

1. (a) Find the mass of the twelfth egg.

   • Total of the 12 masses: \(12\times 59.2=710.4\)
   • Total of the first 11 masses: \(11\times 58.5=643.5\)
   • Twelfth mass: \(710.4-643.5=66.9\) g

2. (b) An egg of mass 52 g is then removed from the 12. Find the mean mass of the other 11 eggs, to 3 significant figures.

   • Total of the other 11: \(710.4-52=658.4\)
   • Mean: \(\frac{658.4}{11}=59.9\) g

The table shows the number of goals scored by a hockey team in each of its matches this season.

1. (a) Find how many matches there were in total.

   • \(3 + 6 + 8 + 6 + 3 + 2 + 1 = 29\)

2. (b) Find how many matches had more than 2 goals.

   • 6 + 3 + 2 + 1 = 12

3. (c) Use your calculator to find the mean.

   • Enter the values in one list and the frequencies in another.
   • From the calculator: x̄ = 2.34

4. (d) Use your calculator to find the standard deviation.

   • From the calculator: σx = 1.51 (use σx, not sx)

A fruit farmer weighs the fruit picked from 20 trees. The table shows the mass of fruit per tree, in kg. The mean mass is 8.55 kg.

1. (a) Write down the total frequency.

   • \(3 + 6 + 5 + 6 = 20\)

2. (b) Find the value of y.

   • \(\frac{2\times3+6\times6+9\times5+6y}{20}=8.55\)
   • \(87+6y=171\)
   • \(6y=84,\ y=14\)

The heights of 80 seedlings in a nursery are recorded in the table.

1. (a) How many seedlings have a height less than 16.5 cm?

   • 8 + 17 + 25 = 50

2. (b) State the modal group.

   • The highest frequency is 25.
   • Modal group: 14.5 ≤ h < 16.5

3. (c) Estimate the mean height.

   • Midpoints: 11.5, 13.5, 15.5, 17.5, 19.5
   • \(\frac{11.5\times8+13.5\times17+15.5\times25+17.5\times20+19.5\times10}{80}=15.7\)

The lengths of phone calls to a shop, in minutes, are grouped in the table. Estimate the mean length of a call.

• Midpoints: 1.5, 4.5, 9.5
• \(\Sigma f x = 14\times1.5+22\times4.5+9\times9.5=205.5\)
• \(\bar{x}\approx\frac{205.5}{45}=4.57\)

The mean time of 12 sprinters from one athletics club is 11.8 seconds for 100 m. The mean time of all 30 sprinters at a meeting is 11.5 seconds.

1. (a) Find the total of the times of all 30 sprinters.

   • \(30\times 11.5=345\)

2. (b) Find the mean time of the other 18 sprinters.

   • Total of the 12 club sprinters: \(12\times 11.8=141.6\)
   • Total of the other 18: \(345-141.6=203.4\)
   • Mean: \(\frac{203.4}{18}=11.3\) seconds

A shampoo maker advertises: "Eight out of ten users prefer our shampoo." The claim comes from a form on the company's website, sent to 6000 users. 150 users filled it in, and 120 of them said yes.

1. (a) Find the percentage of the replies that said yes.

   • \(\frac{120}{150}\times 100=80\%\)

2. (b) Give two reasons why the advertisement may not be reliable.

   • \(\frac{150}{6000}\times 100=2.5\%\): almost everyone asked did not reply, so the replies are self-selected.
   • Users who like the shampoo are more likely to fill in the form, so the sample is biased towards yes.
   • The company wants a good result, so the source is not neutral.

3. (c) Suggest one way to get a more reliable result.

   • Choose users at random from the customer list.
   • Ask a neutral question and follow up the people who do not reply.

The times, in seconds, taken by 15 pupils to solve a puzzle are: 12.4, 18.1, 15.6, 22.3, 13.8, 17.2, 26.0, 14.9, 19.5, 16.3, 21.7, 12.9, 24.1, 18.8, 15.2.

1. (a) Use your calculator to find the exact mean.

   • From the calculator: x̄ = 17.9

2. (b) Complete a grouped frequency table for the classes 10 ≤ t < 15, 15 ≤ t < 20, 20 ≤ t < 25, 25 ≤ t < 30.

   • 10 ≤ t < 15: 4
   • 15 ≤ t < 20: 7
   • 20 ≤ t < 25: 3
   • 25 ≤ t < 30: 1
   • Check: total = 15

3. (c) Use your table to estimate the mean.

   • Midpoints: 12.5, 17.5, 22.5, 27.5
   • \(\frac{12.5\times4+17.5\times7+22.5\times3+27.5\times1}{15}=17.8\)

4. (d) Explain why the answers to (a) and (c) are different.

   • The estimate treats every value in a class as if it were at the midpoint.
   • The actual values are spread across each class, so the two means differ.

The scores of 12 players in a game are: 48, 55, 62, 71, 55, 83, 67, 55, 90, 74, 59, 66.

1. (a) Use your calculator to find the mean.

   • One-variable statistics: x̄ = 65.4

2. (b) Find the median.

   • Ordered list: 48, 55, 55, 55, 59, 62, 66, 67, 71, 74, 83, 90
   • Median = 64

3. (c) Find the mode.

   • 55 appears 3 times, more than any other value.

4. (d) Find the standard deviation σx.

   • From the calculator: σx = 11.9

A gym owner wants to know whether members would like longer opening hours. She asks the first 50 members who arrive on Monday morning.

1. (a) Name the sampling method used.

   • The people were chosen because they were easy to reach: convenience sampling.

2. (b) Give one reason why the sample may be biased.

   • Members who come on Monday morning may be keen early visitors, so they may want longer hours more (or less) than other members; the sample is not representative.

3. (c) Suggest a better way to choose the sample.

   • Use a random sample from the full membership list, or a stratified sample by age group or by visit time.