Question

In the following probability​ distribution, the random variable x represents the number of activities a parent of a 6 th dash to 8 th ​-grade student is involved in. Complete parts​ (a) through​ (f) below.

x 0 1 2 3 4

​P(x) 0.216 0.072 0.322 0.289 0.101

Compute and interpret the mean of the random variable x.

The mean is ___ activities.

​(Type an integer or a​ decimal.)

​(d) Compute the standard deviation of the random variable x.

The standard deviation is _____ activities.

​(Round to one decimal place as​ needed.)

​(e) What is the probability that a randomly selected student has a parent involved in three​ activities?

The probability is _____ ​(Type an integer or a​ decimal.)

​(f) What is the probability that a randomly selected student has a parent involved in three or four​ activities?

The probability is_____ (Type an integer or a​ decimal.)

Answer 1

The mean/expected value is

$E[X]=\displaystyle\sum_x x\,P(X=x)=0P(X=0)+1P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)$

$\implies E[X]=\boxed{1.987}$

d. The standard deviation is the square root of the variance, which itself is

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

We have

$E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=0^2P(X=0)+1^2P(X=1)+2^2P(X=2)+3^2P(X=3)+4^2P(X=4)$

$\implies E[X^2]=5.577$

Then the variance is

$V[X]=5.577-1.987^2\approx1.629$

and so the standard deviation is

$\sqrt{V[X]}\approx\boxed{1.276}$

e. We know this immediately from the table:

$P(X=3)=\boxed{0.289}$

f. A parent can be involved in either 3 or 4 activities, but not simultaneously 3 and 4 activities (i.e. these events are disjoint), so

$P(X=3\text{ or }X=4)=P(X=3)+P(X=4)=\boxed{0.39}$

Answer 2

Applying statistical concepts, we have that:

c) The mean is of 1.987 activities.

d) The standard deviation is of 1.67 activities.

e) The probability is 0.289.

f) The probability is 0.39.

The probability distribution is given by:

$P(X = 0) = 0.216$

$P(X = 1) = 0.072$

$P(X = 2) = 0.322$

$P(X = 3) = 0.289$

$P(X = 4) = 0.101$

Item c:

The mean is given by the sum of the multiplications of each outcome by it's probability, thus:

$E(X) = 0.216(0) + 0.072(1) + 0.322(2) + 0.289(3) + 0.101(4) = 1.987$

The mean is of 1.987 activities.

Item d:

The standard deviation is given by the square root of the sum of the squares of each outcome subtracted by the mean, multiplied by it's probability, thus:

$\sqrt{V(X)} = \sqrt{0.216(0-1.987)^2 + 0.072(1-1.987)^2) + 0.322(2-1.987)^2) + ...}$

$\sqrt{V(X)} = 1.67$

The standard deviation is of 1.67 activities.

Item e:

This probability is P(X = 3), thus 0.289.

Item f:

This probability is:

$p = P(X = 3) + P(X = 4) = 0.289 + 0.101 = 0.39$

A similar problem is given at https://brainly.com/question/14395358