Question

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 $\mu_{\bar X}$ and the variance $\sigma^2_{\bar X}$ of the mean $\bar X$ 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.

Answer 1

Answer:

Step-by-step explanation:

Given is the probability distribution of a random variable X

X 4 5 6 7 Total

P 0.2 0.4 0.3 0.1 1

x*p 0.8 2 1.8 0.7 5.3

x^2*p 3.2 10 10.8 4.9 28.9

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

$Var(x) = 28.9-5.3^2=0.81$

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 = $\frac{0.81}{{36} } \\=0.0225$

c) $P(\bar X$