Amelia decides to analyse the heights of members of her school rowing club. The heights of a random sample of 10 rowers are shown in the table below. | Rower | Height (cm) | |-------|--------------| | Jess | 162 | | Nell | 169 | | Liv | 172 | | Neve | 156 | | Ann | 146 | | Toni | 159 | | Kath | 157 | | Darcy | 160 | | Jen | 160 | 12 (a) Any value more than 2 standard deviations from the mean may be regarded as an outlier. Verify that Ann's height is an outlier. Fully justify your answer. 12 (b) Amelia thinks she may have written down Ann's height incorrectly. If Ann's height were discarded, state a reason what, if any, difference this would make to the mean and standard deviation.

A-Level AQA Maths Pure Proof: Amelia decides to analyse the he

Amelia decides to analyse the heights of members of her school rowing club - AQA - A-Level Maths Pure - Question 12 - 2019 - Paper 3

Question 12

Amelia decides to analyse the heights of members of her school rowing club. The heights of a random sample of 10 rowers are shown in the table below. | Rower | ... show full transcript

Worked Solution & Example Answer:Amelia decides to analyse the heights of members of her school rowing club - AQA - A-Level Maths Pure - Question 12 - 2019 - Paper 3

Step 1

Verify that Ann's height is an outlier.

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Answer

To determine if Ann's height of 146 cm is an outlier, we start by calculating the mean and standard deviation of the heights.

Step 1: Calculate the Mean

The heights are: 162, 169, 172, 156, 146, 159, 157, 160, 160.

Mean ( $ar{x}$ ) is calculated as follows:
$ar{x} = rac{162 + 169 + 172 + 156 + 146 + 159 + 157 + 160 + 160}{9}$
$ar{x} = rac{1601}{9} ext{ which gives } ar{x} ext{ approximately } 160.1$

Step 2: Calculate the Standard Deviation

Next, we compute the standard deviation ( $s$ ):

$s = rac{1}{n-1} imes ext{sqrt}igg( rac{ ext{sum of }(x_i - ar{x})^2}{(n-1)}igg)$

Calculating the variance:

Difference from mean:
- $(162 - 160.1)^2 = 3.61$
- $(169 - 160.1)^2 = 79.21$
- $(172 - 160.1)^2 = 141.61$
- $(156 - 160.1)^2 = 16.81$
- $(146 - 160.1)^2 = 201.61$
- $(159 - 160.1)^2 = 1.21$
- $(157 - 160.1)^2 = 9.61$
- $(160 - 160.1)^2 = 0.01$
- $(160 - 160.1)^2 = 0.01$

Total:
$ext{Total} = 3.61 + 79.21 + 141.61 + 16.81 + 201.61 + 1.21 + 9.61 + 0.01 + 0.01 = 453.9$

Variance = $rac{453.9}{8} ext{ gives } ext{variance } ext{approximately } 56.74$

Standard deviation:
$s ext{ approximately } ext{sqrt}(56.74) ext{ gives } 7.53$

Step 3: Determine Outlier Condition

To check if Ann's height is an outlier, we calculate the cutoff values:

Lower Bound: $ar{x} - 2s = 160.1 - 2(7.53) ext{ which gives } 145.04$ .
Upper Bound: $ar{x} + 2s = 160.1 + 2(7.53) ext{ which gives } 175.16$ .

Since Ann's height of 146 cm is less than the lower bound of 145.04, Ann's height is indeed an outlier.

Step 2

If Ann's height were discarded, state a reason what, if any, difference this would make to the mean and standard deviation.

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Answer

If Ann's height were discarded, the mean would likely increase. This is because Ann's height (146 cm) is below the current mean (approximately 160.1 cm). Removing a lower value would elevate the average.

Effect on Standard Deviation

The standard deviation would generally decrease. This is due to the fact that the overall spread of data decreases when an outlier—in this case, Ann's height—is removed, leading to a tighter cluster of values around the new mean.