Answer :
Answer:
Part a: The value of probability density function is 1/5 or 0.2 for x∈[1,6] and 0 for x∉[1,6].
Part b: The value of mean is 7/2 or 3.5.
Part c: The value of standard deviation is 1.443
Part d: The probability of p(μ-σ≤x≤μ+σ) is 0.5766.
Part e: The probability of x greater than 3.71 is 0.458.
Step-by-step explanation:
As the question is not understandable, the similar question is found online and is attached herewith.
Part a:
The probability density function is given as
[tex]f(x)=\dfrac{1}{c-d}[/tex]
where c and d are limits of the range given here as 6 and 1 respectively so the value of the probability density function is
[tex]f(x)=\dfrac{1}{c-d}\\f(x)=\dfrac{1}{6-1}\\f(x)=\dfrac{1}{5}=0.2[/tex]
The value of probability density function is 1/5 or 0.2 for x∈[1,6] and 0 for x∉[1,6].
Part b:
The mean is given as
[tex]\mu=\dfrac{c+d}{2}[/tex]
By substituting the values
[tex]\mu=\dfrac{c+d}{2}\\\mu=\dfrac{6+1}{2}\\\mu=\dfrac{7}{2}=3.5[/tex]
The value of mean is 7/2 or 3.5.
Part c:
The standard deviation is given as
[tex]\sigma=\dfrac{c-d}{\sqrt{12}}[/tex]
By putting the values in the equation
[tex]\sigma=\dfrac{c-d}{\sqrt{12}}\\\sigma=\dfrac{6-1}{\sqrt{12}}\\\sigma=\dfrac{5}{\sqrt{12}}=1.443[/tex]
The value of standard deviation is 1.443
Part d:
P(μ-σ≤x≤μ+σ) is to be calculated which is given as
P(3.5-1.443≤x≤3.5+1.443)=P(2.057≤x≤4.94)
So the probability is given as
[tex]p(a\leq x\leq b)=\int\limits^a_b {f(x)} \, dx[/tex]
Here a=2.057, b=4.94 and f(x)=0.2 so the equation is as
[tex]p(a\leq x\leq b)=\int\limits^a_b {f(x)} \, dx\\p(2.057\leq x\leq 4.94)=\int\limits^{4.94}_{2.057} {0.2} \, dx\\p(2.057\leq x\leq 4.94)=0.5766[/tex]
The probability of p(μ-σ≤x≤μ+σ) is 0.5766.
Part e:
The probability of x≥3.71 is given as
[tex]p(x\geq 3.71)=\int\limits^{\infty}_{3.71} {f(x)} \, dx \\p(x\geq 3.71)=\int\limits^{6}_{3.71} {f(x)} \, dx +\int\limits^{\infty}_{6} {f(x)} \, dx \\p(x\geq 3.71)=\int\limits^{6}_{3.71} {0.2} \, dx +\int\limits^{\infty}_{6} {0} \, dx \\p(x\geq 3.71)=\int\limits^{6}_{3.71} {0.2} \, dx +0\\p(x\geq 3.71)=0.458[/tex]
So the probability of x greater than 3.71 is 0.458.
From the random variable in this uniform probability question that ranges from 1 to 6,
- PD = 0.2
- mean = 3.5
- sd = 1.4432
a.
probability density = f(x) = 1 / (6-1) = 0.2
b.
mean
u = (6+1) / 2 = 3.5
c.
variance = (1/12) * (6-1)^2
= 2.083
standard deviation =√2.083
= 1.4432
d.
P(u-sigma <=X<= u + sigma)
= 2*1.4432 / 5
= 0.5773
e.
P(X>=3.71)
= (6-3.71) / (6-1)
= 0.458
What is a uniform probabilty?
This is the type of probability has all of its outcomes to be equally likely when they are calculated.
Read more on uniform probability here: https://brainly.com/question/25195809