Suppose is a random variable best described by a uniform probability that ranges from 1 to 6. Compute the following: (a) the probability density function (x)= (b) the mean = 3.5 (c) the standard deviation = (d) P(−≤x≤+)= (e) P(x≥3.71)=

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

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

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

By substituting the values

The value of mean is 7/2 or 3.5.

Part c:

The standard deviation is given as

By putting the values in the equation

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

Here a=2.057, b=4.94 and f(x)=0.2 so the equation is as

The probability of p(μ-σ≤x≤μ+σ) is 0.5766.

Part e:

The probability of x≥3.71 is given as

So the probability of x greater than 3.71 is 0.458.

answer: 59

step-by-step explanation: