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)

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x