Statistics

Have a go first. Then open the working one line at a time.

1. Look first

   Both means are 60. Class P is spread along the scale. Class Q has eight marks on 60 and two far away.

   The means match, so look at how far the marks sit from 60.

2. Distances, Class P

   \(|x-60|\): 7, 5, 3, 2, 1, 1, 2, 3, 5, 7. Sum = 36

   A distance is never negative, so the bars matter.

3. MAD, Class P

   \(\text{MAD}=\frac{36}{10}=3.6\)

   Divide by the number of values.

4. Class Q

   Distances: 18, 0, 0, 0, 0, 0, 0, 0, 0, 18. Sum = 36. \(\text{MAD}=3.6\)

   The same MAD as Class P. MAD cannot tell these classes apart, yet Class Q has two marks very far from the mean.

Have a go first. Then open the working one line at a time.

1. Squared distances, Class P

   \((x-60)^2\): 49, 25, 9, 4, 1, 1, 4, 9, 25, 49. Sum = 176

   Squaring makes a large distance count for much more.

2. Variance and SD, Class P

   \(\sigma^2=\frac{176}{10}=17.6,\quad \sigma=\sqrt{17.6}=4.20\)

   The variance is the mean of the squared distances. The standard deviation is its square root.

3. Class Q

   Squared distances: 324, 0, 0, 0, 0, 0, 0, 0, 0, 324. Sum = 648. \(\sigma^2=64.8,\ \sigma=8.05\)

   Check from the calculator: σx = 8.05.

4. Compare

   Class P: σ = 4.20. Class Q: σ = 8.05.

   Class Q's standard deviation is about twice Class P's. SD sees what MAD missed: Class Q is more spread out.

Have a go first. Then open the working one line at a time.

1. Mean

   \(\bar{x}=\frac{50}{5}=10\)

   Find the mean first.

2. The table

   Deviations x − mean add up to 0. Squares add up to 58.

   If the deviations do not add up to 0, check the mean.

3. Variance

   \(\sigma^2=\frac{58}{5}=11.6\)

   Divide by n.

4. Standard deviation

   \(\sigma=\sqrt{11.6}=3.41\) minutes

   The standard deviation has the same units as the data. Check from the calculator: σx = 3.41.

Have a go first. Then open the working one line at a time.

1. Mean

   \(\bar{x}=\frac{165}{5}=33\)

   The mean of the five times.

2. Standard deviation

   Deviations: −3, 1, −1, 2, 1. Squares add to 16. \(\sigma^2=3.2,\ \sigma=1.789\)

   Square, add, divide by 5, take the root.

3. Distance in SDs

   \(\frac{33-28}{1.789}=2.80\)

   The record is 5 s from the mean. Dividing by σ counts how many standard deviations that is.

4. State

   The record is about 2.80 standard deviations below the mean.

   Far more than 2 standard deviations from the mean: a very unusual time.

Have a go first. Then open the working one line at a time.

1. Before

   From the calculator: x̄ = 15.5, σx = 2.29, median = 15.5, Q1 = 13.5, Q3 = 17.5

   IQR = Q3 − Q1 = 4.

2. After

   From the calculator: x̄ = 36.9, σx = 57.9, median = 15.5, Q1 = 13.5, Q3 = 17.5

   IQR = Q3 − Q1 = 4.

3. Compare

   Mean 15.5 → 36.9. SD 2.29 → 57.9. Median 15.5 → 15.5. IQR 4 → 4.

   One extreme value drags the mean and SD up a long way. The median and IQR do not move.

4. Which to use

   Use median and IQR when the data has outliers or is skewed.

   Use mean and standard deviation when the data is clean and roughly bell-shaped.

Have a go first. Then open the working one line at a time.

1. Write in SDs

   242 = 250 − 2 × 4 and 258 = 250 + 2 × 4

   Both limits are 2 standard deviations from the mean.

2. Part (a)

   About 95% of the values lie within 2 SDs of the mean: 95%

   For bell-shaped data: about 68% within 1 SD, 95% within 2 SDs, 99.7% within 3 SDs.

3. Both tails

   100% − 95% = 5% outside

   The curve is symmetrical, so the two tails are equal.

4. Part (b)

   5% ÷ 2 = 2.5%

   About 2.5% of bags are heavier than 258 g.