Question

for the data shown above, find the following.
do not round any of your answers.

a) find the 5-number summary:
{ 318 × , 225 ✔ , 243 ✔ , 279 ✔ , 318 ✔ }
what units would go on each of your answers above? mg/dl ✔

b) compute the iqr.
put your answer in the first box and the correct units in the second box.
box mg/dl ✔

c) what is the lower fence for this data set?
box mg/dl ✔
what is the upper fence for this data set?
box mg/dl ✔
enter an integer or decimal number more..

Explanation:

Part a) 5 - number summary correction

The 5 - number summary consists of the minimum, \(Q_1\) (first quartile), median (\(Q_2\)), \(Q_3\) (third quartile), and maximum. From the given (with the first value wrong), let's assume the correct minimum is the smallest value. Wait, the correct 5 - number summary should be ordered. Let's re - evaluate. If the correct values (after correction) are: minimum (\(Min\)), \(Q_1\), median (\(Med\)), \(Q_3\), maximum (\(Max\)). From the given, the correct \(Q_1 = 225\), median \(= 243\), \(Q_3=279\), \(Max = 318\). So the minimum was wrong. Let's assume the data set is such that the minimum is the smallest value. Wait, maybe the initial minimum was a typo. Let's proceed with the correct 5 - number summary structure: \(\{Min, Q_1, Med, Q_3, Max\}\). If \(Q_1 = 225\), \(Med = 243\), \(Q_3 = 279\), \(Max=318\), then we need to find the correct minimum. But since the user's initial minimum was wrong, but for the sake of solving part b, c, we can use the given \(Q_1 = 225\), \(Q_3 = 279\), \(Min\) (let's assume the correct minimum is, say, if the data set is ordered, maybe the first value was a mistake. But let's proceed with the given \(Q_1 = 225\), \(Q_3 = 279\), \(Max = 318\), median \(= 243\).

Part b) Compute the IQR

Step 1: Recall the formula for IQR

The Inter - Quartile Range (IQR) is calculated as \(IQR=Q_3 - Q_1\)

Step 2: Substitute the values of \(Q_3\) and \(Q_1\)

We know that \(Q_1 = 225\) and \(Q_3=279\). So, \(IQR = 279-225\)
\(IQR = 54\)

Part c) Lower and Upper Fences

The formula for the lower fence (\(LF\)) is \(LF=Q_1 - 1.5\times IQR\) and the formula for the upper fence (\(UF\)) is \(UF = Q_3+1.5\times IQR\)

Step 1: Calculate the lower fence

We know \(Q_1 = 225\) and \(IQR = 54\). Then, \(1.5\times IQR=1.5\times54 = 81\)
\(LF=225 - 81\)
\(LF = 144\)

Step 2: Calculate the upper fence

We know \(Q_3 = 279\) and \(IQR = 54\). Then, \(1.5\times IQR = 81\)
\(UF=279 + 81\)
\(UF = 360\)

Final Answers

Part b)

Step1: Recall IQR formula

\(IQR = Q_3 - Q_1\)

Step2: Substitute \(Q_3 = 279\), \(Q_1 = 225\)

\(IQR=279 - 225 = 54\)

Step1: Lower fence formula

\(LF=Q_1-1.5\times IQR\)
\(1.5\times IQR = 1.5\times54 = 81\), \(LF = 225-81=144\)

Step2: Upper fence formula

\(UF = Q_3 + 1.5\times IQR\)
\(1.5\times IQR=81\), \(UF=279 + 81 = 360\)

Answer:

54 mg/dL

Part c)