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a bakery sells 10 types of sandwiches. here are their calorie amounts. …

Question

a bakery sells 10 types of sandwiches. here are their calorie amounts. 282, 597, 597, 605, 611, 613, 614, 615, 617, 618 complete the parts below to identify any outliers. (a) let $q_1$ be the lower quartile and $q_3$ be the upper quartile of the data - set. find $q_1$ and $q_3$ for the data set. $q_1 = square$ $q_3=square$ (b) find the interquartile range (iqr) of the data set. $iqr=square$ (c) calculate a lower boundary using $q_1 - 1.5cdot iqr$. calculate an upper boundary using $q_3 + 1.5cdot iqr$. (note that $1.5cdot iqr$ means 1.5 times the iqr.) lower boundary: $square$ upper boundary: $square$ (d) any values less than the lower boundary or greater than the upper boundary are considered outliers. identify all the outliers of the data set. if there is more than one outlier, separate them with commas. if there are no outliers, click
one\. outliers: $square$

Explanation:

Step1: Sort the data

$282,597,597,605,611,613,614,615,617,618$

Step2: Find the median (Q2)

Since $n = 10$ (even), $Q2=\frac{611 + 613}{2}=612$

Step3: Find Q1

The lower - half of the data is $282,597,597,605,611$. Since $n = 5$ (odd), $Q1 = 597$

Step4: Find Q3

The upper - half of the data is $613,614,615,617,618$. Since $n = 5$ (odd), $Q3=615$

Step5: Calculate IQR

$IQR=Q3 - Q1=615 - 597 = 18$

Step6: Calculate lower boundary

Lower boundary $=Q1-1.5\times IQR=597-1.5\times18=597 - 27=570$

Step7: Calculate upper boundary

Upper boundary $=Q3 + 1.5\times IQR=615+1.5\times18=615 + 27=642$

Step8: Identify outliers

The value $282<570$, so the outlier is $282$

Answer:

(a) $Q1 = 597$
$Q3 = 615$
(b) $IQR = 18$
(c) Lower boundary: $570$
Upper boundary: $642$
(d) Outliers: $282$