The answer to the problem
1 answer:
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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.
Step-by-step explanation:
We start with Left hand side
We know that csc(x) = 1/ sin(x)
So csc(2x) is replaced by 1/sin(2x)
Also we use identity
sin(2x) = 2 sin(x) cos(x)
4 divide by 2 is 2
Now we multiply top and bottom by sin(x) because we need tan(x) in our answer
We know that sinx/ cosx = tan(x)
Also 1/ sin(x)= csc(x)
so it becomes 2csc^2(x) tan(x) , Right hand side
Hence verified