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WARRIOR [948]
3 years ago
10

he data represents the daily rainfall​ (in inches) for one month. Construct a frequency distribution beginning with a lower clas

s limit of 0.000.00 and use a class width of 0.200.20. Does the frequency distribution appear to be roughly a normal​ distribution? 0.310.31 0 0 0 0.190.19 0 0.150.15 0 0.010.01 0.190.19 0.530.53 0 0

Mathematics
1 answer:
Licemer1 [7]3 years ago
4 0

Answer:

It is not normally distributed as it has it main concentration in only one side.

Step-by-step explanation:

So, we are given that the class width is equal to 0.2. Thus we will have that the first class is 0.00 - 0.20, second class is 0.20 - 0.40 and so on(that is 0.2 difference).

So, let us begin the groupings into their different classes, shall we?

Data given:

0.31 0.31 0 0 0 0.19 0.19 0 0.150.15 0 0.01 0.01 0.19 0.19 0.53 0.53 0 0.

(1). 0.00 - 0.20: there are 15 values that falls into this category. That is 0 0 0 0.19 0.19 0 0.15 0.15 0 0.01 0.01 0.19 0.19 0 0.

(2). 0.20 - 0.40: there are 2 values that falls into this category. That is 0.31 0.31

(3). 0.4 - 0.6 : there are 2 values that falls into this category.

(4). 0.6 - 0.8: there 0 values that falls into this category. That is 0.53 0.53.

Class interval            frequency.

0.00 - 0.20.                   15.

0.20 - 0.40.                    2.

0.4 - 0.6.                        2.

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2 years ago
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Answer:

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Step-by-step explanation:

If there are two points (x1,y1) and (x2,y2) on the coordinate plane

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_________________________________________________

in the problem midpoint is m(c,n)

one point is Q(h,s)

let other point be P(x,y)

By using midpoint formula given above

for point P(x,y) and Q(h,s)

midpoint = (x+h)/2, (y+s)/2

also midpoint is m(c,n)

comparing m(c,n) with (x+h)/2, (y+s)/2

c = x+h/2

=> 2c = x+h

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n = (y+s)/2

=> 2n = y+s

=> 2n - s = y

Thus, second endpoint is P(x,y) = P(2c - h, 2n - s )

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3 years ago
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