(-3) + (+1) = steps of solving
2 answers:
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Answer:
1. {18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }
2a) md= 24 b) Q1=20.5 c) Q3= 30 3) Q3-Q1 =9.5 b) 19/48
Step-by-step explanation:
To answer this question the 1st and the 2nd we need to order the data entries. So from ordering from the lowest to the highest value:
1. {18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }
2. There are 25 entries.
{18, 18 , 19 , 19 , 19 , 20 , 21 , 21 , 21 , 21 , 23 , 24 , 24 , 26 , 27 , 27 , 29 , 30 , 30 , 30 , 33 , 33, 34 , 35 , 38 }
In odd quantities of observations, the Median equally separates it two parts.
md=24
b) To find out the 1st quartile, we can use this way:
Then 6.5 is between the 6th and 7th position. Let's find the mean of them, now:
c) Similarly toThe Third Quartile or Upper Quartile
The 19th position and 20th position average is:
3)
a) To find the Interquartile Range, we just need to find out the difference of the upper quartile and the lower one:Q_3-Q_1[/tex]
(30-20.5)=9.5
b) Interquartile Ratio is given by the quotient of the Interquartile Range over the Median
4) Since the Relative frequency Histogram asked is a one with 7 classes. Let's calculate how many values.
k=1+3.32logn
7=1+3.32logn
6=3.32logn
n≈66
Each class must have an interval of 10 ages, for (91-18)/7≈ 10. Notice the orange line intercepts the midpoint of each interval.
