6. Suppose we have the cdf of a continuous distribution with random variable X given by the function for x <0 F(x) = (x5 – 2r3 + 2x) for 0 2 What is the probability that X is between 1 and 2? 3 А. 10 19 В. 20 1 C. 20 D. 1 E. None of the above 7. For pdf f(x), which is nonzero on the interval [a, b], the corresponding cdf on the interval [a, b) is given by F(x) = | 5(t)dt A. True B. False 8. For type I error a and type II error B, a = 1 – B. Α. True B. False
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![6.
Suppose we have the cdf of a continuous distribution with random variable
X given by the function
for x <0
F(x) =
(x³ – 2x3 + 2x) for 0 <x < 2
20
for x > 2
What is the probability that X is between 1 and 2?
3
A.
10
19
1
C.
20
D. 1
E. None of the above
20
7.
For pdf f(x), which is nonzero on the interval [a, b], the corresponding cdf on
the interval [a, b] is given by
F(æ) = | f(t)dt
A. True
B. False
8.
For type I error a and type II error 3, a =1- B.
A. True
B. False
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Calculation Part:
6)
The cdf of X is given by
F(X)=0, for x<0
F(X)={1/20(x5-2x3+2x)}, for 0≤X≤2
F(X)=1, for x>1
To find Probability that X lies between 1 and 2:
P(1<X<2)=F(2)-F(1)
Note: P(a<X<b)=F(b)-F(a)
now we have to find F(2) and F(1) separately
F(2)=1/20(25-2*23+2*2)
F(2)=1/20(32-16+4)
F(2)=1/20(20)
F(2)=1
F(1)=1/20(15-2*13+2*1)
F(1)=1/20(1-2+2)
F(1)=1/20(1)
F(1)=1/20
substitute F(1) and F(2) in above to get P(1<X<2)
P(1<X<2)=F(2)-F(1)
P(1<X<2)=1-(1/20)
P(1<X<2)=19/20
Result:
The Probability that X lies between 1 and 2 is 19/20
Hence, the option b is correct.
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