Respuesta :
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
[tex]Var(x) = 28.9-5.3^2=0.81[/tex]
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 = [tex]\frac{0.81}{{36} } \\=0.0225[/tex]
c) [tex]P(\bar X <5.5)\\= P(Z<\frac{5.5-5.3}{\sqrt{0.0225} } \\= P(Z<1.33)\\=0.908[/tex]
The mean and the variance from the data about the cherries will be 5.3 and 0.81.
How to calculate the mean and variance?
The required mean from the information given will be:
= (4 × 0.2) + (5 × 0.4) + (6 × 0.3) + (7 × 0.1)
= 0.8 + 2.0 + 1.8 + 0.7
= 5.3
The variance will be calculated thus:
= (4² × 0.2) + (5² × 0.4) + (6² × 0.3) + (7² × 0.1) - (5.3)²
= 3.2 + 10 + 10.8 + 4.9 - (5.3)²
= 0.81
The variance is 0.81.
The probability that the average number of cherries in 36 cherry puffs will be less than 5.5 will be:
= P(x = 4( + P(x = 5)
= 0.2 + 0.4
= 0.6
Learn more about mean on:
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