Respuesta :
(a) The values for a is 6 and b is 10.
(b) The mean of this uniform distribution is 8.
(c) The standard deviation is 1.1094
(d) Total area below any probability distribution curve is always 1.
(e) The probability that the random variable is more than 7 is 0.75.
(f) The probability that the random variable is between 7 and 9 is 0.5.
(g) The probability that the random variable is equal to 7.91 is 0.4775.
A uniform distribution is defined over the interval from 6 to 10.
a. We have to find the values for a and b.
X = uniform(α = 6, β = 10)
So a = 6, b = 10
b. We have to find the mean of this uniform distribution.
mean = a+b/2
mean = 6+10/2
mean = 16/2
mean = 8
c. We have to find the standard deviation.
Variance = 1/12 (b-a)^2
Variance = 1/12 (10-6)^2
Variance = 1/12 (4)^2
Variance = 1/12 *16
Variance = 1.2308
Standard deviation = √Variance
Standard deviation = √1.2308
Standard deviation = 1.1094
d. We have to check the total area is 1.00 or not.
Total area below any probability distribution curve is always 1.
e. We have to find the probability that the random variable is more than 7.
X = uniform(α = 6, β = 10)
P(X>7) = 1-P(X<7)
P(X>7) = 1-{(7-a)/(b-a)}
P(X>7) = 1-{(7-6)/(10-6)}
P(X>7) = 1- 1/4
P(X>7) = 0.75
f. We have to find the probability that the random variable is between 7 and 9.
X = uniform(α = 6, β = 10)
P(7< X <9) = 9-7/b-a
P(7< X <9) = 9-7/10-6
P(7< X <9) = 2/4
P(7< X <9) = 0.5
g. We have to find the probability that the random variable is equal to 7.91.
X = uniform(α = 6, β = 10)
P(X=7.91) = {(7.91-a)/(b-a)}
P(X=7.91) = {(7.91-6)/(10-6)}
P(X=7.91) = 1.91/4
P(X=7.91) = 0.4775
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