The random variable x, representing the number of cherries in a cherry puff, has the following probability distribution: x: 4 - 5 - 6 - 7 p(x=x): 0.2 - 0.4 - 0.3 - 0.1 (a) find the mean μ and the variance σ² of x.(b) find the mean and the variance of the mean for random samples of 36 cherry puffs.(c) find the probability that the average number of cherries in 36 cherry puffs will be less than 5.5.

Step-by-step explanation:

Given is the probability distribution of a random variable X

X4567Total

P0.20.40.30.11

x*p0.821.80.75.3

x^2*p3.21010.84.928.9

a) E(X) = Mean of X = sum of xp = 5.3

Std dev = square root of variance = 0.9

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b) For sample mean we have

Mean = 5.3

Variance = var(x)/n =

c)

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.