Question

part 1 of 2
points: 0.83 of 1
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listed below are amounts of strontium - 90 (in millibecquerels, or mbq) in a simple random sample of baby teeth obtained from residents in a region born after 1979. use the given data
to construct a boxplot and identify the 5 - number summary.
122 125 127 132 133 135 138 141 144 147
149 151 151 153 154 159 160 162 165 172
the 5 - number summary is □,□,□,□, and □, all in mbq.
(use ascending order. type integers or decimals. do not round.)

Explanation:

Step1: Find the minimum value

The minimum value is the smallest number in the data set. Looking at the data: \(122, 125, 127, 132, 133, 135, 138, 141, 144, 147, 149, 151, 151, 153, 154, 159, 160, 162, 165, 172\), the minimum is \(122\).

Step2: Find the first quartile (\(Q_1\))

The formula for the position of \(Q_1\) is \(i=\frac{n + 1}{4}\), where \(n = 20\) (the number of data points). So \(i=\frac{20+1}{4}=5.25\).
The first quartile is \(Q_1=x_5+(x_6 - x_5)\times0.25\). \(x_5 = 133\), \(x_6=135\). Then \(Q_1=133+(135 - 133)\times0.25=133 + 0.5=133.5\).

Step3: Find the median (\(Q_2\))

The formula for the position of the median is \(i=\frac{n+1}{2}\). For \(n = 20\), \(i=\frac{20 + 1}{2}=10.5\).
The median is \(Q_2=x_{10}+(x_{11}-x_{10})\times0.5\). \(x_{10}=147\), \(x_{11}=149\). So \(Q_2=147+(149 - 147)\times0.5=147 + 1=148\).

Step4: Find the third quartile (\(Q_3\))

The formula for the position of \(Q_3\) is \(i=\frac{3(n + 1)}{4}\). For \(n = 20\), \(i=\frac{3\times(20 + 1)}{4}=15.75\).
The third quartile is \(Q_3=x_{15}+(x_{16}-x_{15})\times0.75\). \(x_{15}=154\), \(x_{16}=159\). Then \(Q_3=154+(159 - 154)\times0.75=154+3.75 = 157.75\).

Step5: Find the maximum value

The maximum value is the largest number in the data set. The maximum is \(172\).

Answer:

\(122\), \(133.5\), \(148\), \(157.75\), \(172\)