As cars passed a checkpoint, the following speeds were clocked and recorded. speed (mph): 55 62 61 54 68 72 59 61 70 what are the minimum, first quartile, median, third quartile, and maximum of the data set? show your work.

We have >  55, 62, 61, 54, 68, 72, 59 ,61, 70  

In total 9 numbers

step 1

order  the numbers  from smallest to largest

54,55,59,61,61,62,68,70,72

1) What is the minimum of the data set ?

The lowest number is the minimum > 54

2) What is the median of the data set ?

Find the number that is in the middle  

54,55,59,61,61,62,68,70,72

The middle number is 61. That is the median.  

Find the middle of each half. Since these numbers are between 2 numbers, you average the numbers (add then divide by 2).  

55+59=114/2=57 This is the first quartile.  

68+70=138/2=69 This is the third quartile

3) What is the first quartile of the data set ?

the answer is 57

4) What is the third quartile of the data set ?

the answer is 69

5) What is the maximum of the data set ?

The highest is the maximum > 72

Minimum = 54

Maximum = 72

Median = 61

First quartile = 57

Third quartile = 69

Step-by-step explanation:

Data : 55 62 61 54 68 72 59 61 70

Arrange the data is ascending order

54 , 55, 59, 61,61 ,62 , 68,70 ,72

So, minimum = 54

Maximum = 72

Median is the mid value .

There are nine observations.

So, 5th observation is the median

So, Median = 61

First quartile = It is the median of data of the left side data of the median .

Left side data : 54 , 55, 59, 61

There are four observations

Median =

Median =

Median =

Median =

Median =

First quartile = 57

Third quartile = It is the median of data of the right side data of the median .

Right side data : 62 , 68,70 ,72

There are four observations

Median =

Median =

Median =

Median =

Median =

Third quartile = 69

Hence minimum is 54 , maximum is 72 ,median is 61 , first quartile is 57 and third quartile is 69

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