题目
Problem
A researcher is studying the birth weights of babies. A random sample of 98 babies was taken and their birth weights, w w w
Birth weight (w w w Frequency (f f f Birth weight midpoint (x x x 1.50 ≤ w < 2.50 1.50\leq w<2.50 1.50 ≤ w < 2.50 16 2.00 2.50 ≤ w < 3.00 2.50\leq w<3.00 2.50 ≤ w < 3.00 24 2.75 3.00 ≤ w < 3.50 3.00\leq w<3.50 3.00 ≤ w < 3.50 32 3.25 3.50 ≤ w < 4.00 3.50\leq w<4.00 3.50 ≤ w < 4.00 14 3.75 4.00 ≤ w < 5.50 4.00\leq w<5.50 4.00 ≤ w < 5.50 12 4.75
(You may use ∑ f x = 311.5 \sum fx=311.5 ∑ f x = 311.5 ∑ f x 2 = 1051.125 \sum fx^2=1051.125 ∑ f x 2 = 1051.125
A histogram is drawn to represent these data.
The bar representing the birth weight 1.50 ≤ w < 2.50 1.50\leq w<2.50 1.50 ≤ w < 2.50
(a) Calculate the width and height of the bar representing birth weight 3.50 ≤ w < 4.00 3.50\leq w<4.00 3.50 ≤ 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}} skewness = standard deviation 3 ( mean − median ) 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 < 3.50 3.00\leq w<3.50 3.00 ≤ w < 3.50
The frequency for 3.00 ≤ w < 3.25 3.00\leq w<3.25 3.00 ≤ w < 3.25 3.25 ≤ w < 3.50 3.25\leq w<3.50 3.25 ≤ w < 3.50
(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)
题目中文翻译
一名研究人员正在研究婴儿的出生体重。随机抽取了 98 名婴儿，他们的出生体重 w w w
出生体重（w w w 频数（f f f 出生体重组中值（x x x 1.50 ≤ w < 2.50 1.50\leq w<2.50 1.50 ≤ w < 2.50 16 2.00 2.50 ≤ w < 3.00 2.50\leq w<3.00 2.50 ≤ w < 3.00 24 2.75 3.00 ≤ w < 3.50 3.00\leq w<3.50 3.00 ≤ w < 3.50 32 3.25 3.50 ≤ w < 4.00 3.50\leq w<4.00 3.50 ≤ w < 4.00 14 3.75 4.00 ≤ w < 5.50 4.00\leq w<5.50 4.00 ≤ w < 5.50 12 4.75
（可使用 ∑ f x = 311.5 \sum fx=311.5 ∑ f x = 311.5 ∑ f x 2 = 1051.125 \sum fx^2=1051.125 ∑ f x 2 = 1051.125
现用直方图表示这些数据。区间 1.50 ≤ w < 2.50 1.50\leq w<2.50 1.50 ≤ w < 2.50
(a) 计算区间 3.50 ≤ w < 4.00 3.50\leq w<4.00 3.50 ≤ w < 4.00
(b) 用线性插值估计这 98 名婴儿出生体重的下四分位数。
研究人员估计中位数为 3.14 kg，上四分位数为 3.55 kg。
(c) 利用中位数与四分位数描述这些数据的偏态。
(d) 估计：(i) 平均出生体重；(ii) 出生体重的标准差。
(e) 使用公式
skewness = 3 ( mean − median ) standard deviation \text{skewness}=\frac{3(\text{mean}-\text{median})}{\text{standard deviation}} skewness = standard deviation 3 ( mean − median ) 估计这些数据的偏度，答案取 2 位有效数字。
研究人员读到出生体重应近似服从正态分布，因此决定把区间 3.00 ≤ w < 3.50 3.00\leq w<3.50 3.00 ≤ w < 3.50 3.00 ≤ w < 3.25 3.00\leq w<3.25 3.00 ≤ w < 3.25 3.25 ≤ w < 3.50 3.25\leq w<3.50 3.25 ≤ w < 3.50
(f) (i) 说明这一改变会怎样影响中位数的估计值，并给出理由。
(ii) 不作进一步计算，说明这一改变会怎样影响平均数的估计值，并给出理由。
解答
(a)
解法一
思路
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直方图面积代表频数。第一组频数 16，对应面积 1 × 4 = 4 1\times4=4 1 × 4 = 4 14 / 16 14/16 14/16
答题过程
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The class 1.50 ≤ w < 2.50 1.50\leq w<2.50 1.50 ≤ w < 2.50 1 1 1 1 1 1
So the class 3.50 ≤ w < 4.00 3.50\leq w<4.00 3.50 ≤ w < 4.00 0.5 0.5 0.5
0.5 cm . \begin{align*}
0.5\text{ cm}.
\end{align*} 0.5 cm . The first bar has area
1 × 4 = 4. \begin{align*}
1\times4=4.
\end{align*} 1 × 4 = 4. This represents frequency 16 16 16 14 14 14
4 × 14 16 = 3.5. \begin{align*}
4\times\frac{14}{16}=3.5.
\end{align*} 4 × 16 14 = 3.5. Hence its height is
3.5 0.5 = 7 cm . \begin{align*}
\frac{3.5}{0.5}=7\text{ cm}.
\end{align*} 0.5 3.5 = 7 cm .
(b)
解法一
思路
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下四分位数位置是第 98 / 4 = 24.5 98/4=24.5 98/4 = 24.5 2.50 ≤ w < 3.00 2.50\leq w<3.00 2.50 ≤ w < 3.00
答题过程
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The lower quartile is the 24.5 24.5 24.5
The cumulative frequency before 2.50 ≤ w < 3.00 2.50\leq w<3.00 2.50 ≤ w < 3.00 16 16 16
So the lower quartile is
2.50 + 24.5 − 16 24 × 0.50. \begin{align*}
2.50+\frac{24.5-16}{24}\times0.50.
\end{align*} 2.50 + 24 24.5 − 16 × 0.50. Thus
Q 1 = 2.677 … . \begin{align*}
Q_1=2.677\ldots.
\end{align*} Q 1 = 2.677 … . So the lower quartile is approximately
2.68 kg . \begin{align*}
2.68\text{ kg}.
\end{align*} 2.68 kg .
(c)
解法一
思路
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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
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Q 2 − Q 1 = 3.14 − 2.68 = 0.46. \begin{align*}
Q_2-Q_1=3.14-2.68=0.46.
\end{align*} Q 2 − Q 1 = 3.14 − 2.68 = 0.46. Also,
Q 3 − Q 2 = 3.55 − 3.14 = 0.41. \begin{align*}
Q_3-Q_2=3.55-3.14=0.41.
\end{align*} Q 3 − Q 2 = 3.55 − 3.14 = 0.41. Since
Q 2 − Q 1 > Q 3 − Q 2 , \begin{align*}
Q_2-Q_1>Q_3-Q_2,
\end{align*} Q 2 − Q 1 > Q 3 − Q 2 , the data are negatively skewed.
(d)
(i)
思路
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题目已给 ∑ f x \sum fx ∑ f x
答题过程
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The estimated mean is
w ˉ = ∑ f x 98 = 311.5 98 = 3.17857 … . \begin{align*}
\bar w=\frac{\sum fx}{98}=\frac{311.5}{98}=3.17857\ldots.
\end{align*} w ˉ = 98 ∑ f x = 98 311.5 = 3.17857 … . So
w ˉ = 3.18 kg . \begin{align*}
\bar w=3.18\text{ kg}.
\end{align*} w ˉ = 3.18 kg .
(ii)
思路
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使用题目给出的 ∑ f x 2 \sum fx^2 ∑ f x 2
答题过程
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The estimated standard deviation is
σ = ∑ f x 2 98 − w ˉ 2 = 1051.125 98 − ( 3.17857 … ) 2 = 0.78895 … . \begin{align*}
\sigma
=&\,\sqrt{\frac{\sum fx^2}{98}-\bar w^2}\\
=&\,\sqrt{\frac{1051.125}{98}-(3.17857\ldots)^2}\\
=&\,0.78895\ldots.
\end{align*} σ = = = 98 ∑ f x 2 − w ˉ 2 98 1051.125 − ( 3.17857 … ) 2 0.78895 … . So
σ = 0.789 kg . \begin{align*}
\sigma=0.789\text{ kg}.
\end{align*} σ = 0.789 kg .
(e)
解法一
思路
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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
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skewness = 3 ( 3.17857 … − 3.14 ) 0.78895 … = 0.1466 … . \begin{align*}
\text{skewness}
=&\,\frac{3(3.17857\ldots-3.14)}{0.78895\ldots}\\
=&\,0.1466\ldots.
\end{align*} skewness = = 0.78895 … 3 ( 3.17857 … − 3.14 ) 0.1466 … . To 2 significant figures,
skewness = 0.15. \begin{align*}
\text{skewness}=0.15.
\end{align*} skewness = 0.15.
(f)
(i)
思路
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第 49 个数据现在落在 3.25 ≤ w < 3.50 3.25\leq w<3.50 3.25 ≤ w < 3.50
答题过程
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The median will increase, because the 49th value now lies in the class 3.25 ≤ w < 3.50 3.25\leq w<3.50 3.25 ≤ w < 3.50
(ii)
思路
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拆组后，较高半组的频数 23 大于较低半组的频数 9，因此用新组中值估算时，∑ f x \sum fx ∑ f x
答题过程
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The mean will increase, because more values are in the higher half of the split class, so the estimated value of ∑ f x \sum fx ∑ f x
