How to solve 4a+8>11a+15and 13-14a<13-3a
1 answer:
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Answer:
(a) The median estimate is 450.
(b) The first and third quartiles are 300 and 750 respectively.
(c) The interquartile range is 450.
(d) The particular value which is less than 150 or it is more than 1200, is considered as an outlier.
(e) Yes, there is one outlier in our data which is the value of 1,500.
(f) The distribution is positively skewed.
Step-by-step explanation:
We are given that the box plot shows the undergraduate in-state tuition per credit hour at four-year public colleges. 0, 300, 600, 900, 1,200, 1,500.
(a) As we can see in the box plot that the middlemost data lies between 300 and 600, that means;
Median is given by the average of the two numbers, i.e;
Median =
= = 450
So, the median estimate is 450.
(b) As we know that the first quartile represents 25% of the data values and the third quartile represents 75% of the data values.
So, from the box plot given; we can observe that the first quartile, = 300 and the third quartile lies between the values of 600 and 900, i.e;
Third quartile, =
= = 750
So, the first and third quartiles are 300 and 750 respectively.
(c) The interquartile range is given by the following formula;
Interquartile range = Third quartile - First quartile
=
= 750 - 300 = 450
(d) From the box plot; it is clear that if the particular value is less than 150 or it is more than 1200, then it is considered as an outlier.
(e) Yes, there is one outlier in our data which is the value of 1,500.
(f) The distribution is positively skewed because the more values are on the right side of the curve and it skewed towards the right.