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IAL 2020 Jan Q4

A Level / Edexcel / S1

IAL 2020 Jan Paper · Question 4

题目

Problem

A researcher is studying the birth weights of babies. A random sample of 98 babies was taken and their birth weights, ww kg, are summarised in the table below.

Birth weight (ww kg)Frequency (ff)Birth weight midpoint (xx)
1.50≤w<2.501.50\leq w<2.50162.00
2.50≤w<3.002.50\leq w<3.00242.75
3.00≤w<3.503.00\leq w<3.50323.25
3.50≤w<4.003.50\leq w<4.00143.75
4.00≤w<5.504.00\leq w<5.50124.75

(You may use ∑fx=311.5\sum fx=311.5 and ∑fx2=1051.125\sum fx^2=1051.125)

A histogram is drawn to represent these data.

The bar representing the birth weight 1.50≤w<2.501.50\leq w<2.50 has a width of 1 cm and a height of 4 cm.

(a) Calculate the width and height of the bar representing birth weight 3.50≤w<4.003.50\leq w<4.00

(3)

(b) Use linear interpolation to estimate the lower quartile of the birth weights of the 98 babies.

(2)

The researcher estimated the median to be 3.14 kg and the upper quartile to be 3.55 kg.

(c) Use the median and quartiles to describe the skewness of these data.

(2)

(d) Find an estimate for

(i) the mean birth weight

(ii) the standard deviation of the birth weights.

(3)

(e) Use the formula

skewness=3(mean−median)standard deviation\text{skewness}=\frac{3(\text{mean}-\text{median})}{\text{standard deviation}}

to estimate a value for the skewness of these data. Give your answer to 2 significant figures.

(2)

The researcher read that birth weights should be approximately normally distributed and decides to split the class 3.00≤w<3.503.00\leq w<3.50

The frequency for 3.00≤w<3.253.00\leq w<3.25 is 9 and the frequency for 3.25≤w<3.503.25\leq w<3.50 is 23

(f) (i) State, giving a reason, what the effect would be on the estimate of the median.

(ii) Without carrying out any further calculations state, giving a reason, what the effect of this change would be on the estimate of the mean.

(2)

解答

(a)

解法一

思路

展开

直方图面积代表频数。第一组频数 16，对应面积 1×4=41\times4=4，所以频数 14 对应面积 14/1614/16 倍。目标组的真实组距是 0.5 kg，图上宽度也按同一比例变成 0.5 cm。

答题过程

展开

The class 1.50≤w<2.501.50\leq w<2.50 has class width 11 kg and is drawn with width 11 cm.

So the class 3.50≤w<4.003.50\leq w<4.00, which has class width 0.50.5 kg, has drawn width

0.5 cm.\begin{align*} 0.5\text{ cm}. \end{align*}

The first bar has area

1×4=4.\begin{align*} 1\times4=4. \end{align*}

This represents frequency 1616. The bar for frequency 1414 therefore has area

4×1416=3.5.\begin{align*} 4\times\frac{14}{16}=3.5. \end{align*}

Hence its height is

3.50.5=7 cm.\begin{align*} \frac{3.5}{0.5}=7\text{ cm}. \end{align*}

(b)

解法一

思路

展开

下四分位数位置是第 98/4=24.598/4=24.5 个数据，落在 2.50≤w<3.002.50\leq w<3.00 这一组。用线性插值。

答题过程

展开

The lower quartile is the 24.524.5th value.

The cumulative frequency before 2.50≤w<3.002.50\leq w<3.00 is 1616.

So the lower quartile is

2.50+24.5−1624×0.50.\begin{align*} 2.50+\frac{24.5-16}{24}\times0.50. \end{align*}

Thus

Q1=2.677….\begin{align*} Q_1=2.677\ldots. \end{align*}

So the lower quartile is approximately

2.68 kg.\begin{align*} 2.68\text{ kg}. \end{align*}

(c)

解法一

思路

展开

比较中位数到上下四分位数的距离。左边距离更大，说明左尾较长，是负偏态。

答题过程

展开 Q2−Q1=3.14−2.68=0.46.\begin{align*} Q_2-Q_1=3.14-2.68=0.46. \end{align*}

Also,

Q3−Q2=3.55−3.14=0.41.\begin{align*} Q_3-Q_2=3.55-3.14=0.41. \end{align*}

Since

Q2−Q1>Q3−Q2,\begin{align*} Q_2-Q_1>Q_3-Q_2, \end{align*}

the data are negatively skewed.

(d)

解法一

思路

展开

题目已给 ∑fx\sum fx 和 ∑fx2\sum fx^2，直接用分组数据平均数和标准差公式。

答题过程

展开

The estimated mean is

wˉ=∑fx98=311.598=3.17857….\begin{align*} \bar w=\frac{\sum fx}{98}=\frac{311.5}{98}=3.17857\ldots. \end{align*}

So

wˉ=3.18 kg.\begin{align*} \bar w=3.18\text{ kg}. \end{align*}

The estimated standard deviation is

σ= ∑fx298−wˉ2= 1051.12598−(3.17857…)2= 0.78895….\begin{aligned} \sigma =&\,\sqrt{\frac{\sum fx^2}{98}-\bar w^2}\\ =&\,\sqrt{\frac{1051.125}{98}-(3.17857\ldots)^2}\\ =&\,0.78895\ldots. \end{aligned}

So

σ=0.789 kg.\begin{align*} \sigma=0.789\text{ kg}. \end{align*}

(e)

解法一

思路

展开

直接把 (d) 的平均数、标准差以及题目给的 median 代入公式。

答题过程

展开 skewness= 3(3.17857…−3.14)0.78895…= 0.1466….\begin{aligned} \text{skewness} =&\,\frac{3(3.17857\ldots-3.14)}{0.78895\ldots}\\ =&\,0.1466\ldots. \end{aligned}

To 2 significant figures,

skewness=0.15.\begin{align*} \text{skewness}=0.15. \end{align*}

(f)

解法一

思路

展开

分裂后，中位数所在的组更集中在 3.25≤w<3.503.25\leq w<3.50，所以中位数估计会上升。均值方面，把原本 32 个数据都放在 3.25，现在其中 9 个放到更低的 3.125，23 个放到更高的 3.375；高的一边更多，所以均值会上升。

答题过程

展开

The median will increase, because the median value will now lie in the class 3.25≤w<3.503.25\leq w<3.50.

The mean will also increase, because more of the values in the split class are placed in the higher half of the interval.