Question

It is given that X-N(1.5,3.2). Find the probability that a randomly chosen value of X is less than −2.4

Answer 1

Probability that a randomly chosen value of X is less than 2.4 is 0.69497.

Step-by-step explanation:

We are given the following normal distribution;

X  ~ Normal($\mu=1.5,\sigma^{2} =3.2$)

The z score probability distribution for normal distribution is given by;

                         Z =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean = 1.5

            $\sigma$ = standard deviation = $\sqrt{3.2}$ = 1.78

Now, probability that a randomly chosen value of X is less than 2.4 is given by = P(X < 2.4)

            P(X < 2.4) = P( $\frac{X-\mu}{\sigma}$ < $\frac{2.4-1.5}{1.78}$ ) = P(Z < 0.51) = 0.69497

The above probability is calculated by looking at the value of x = 0.51 in the z table which has an area of 0.69497.

Answer 2

Answer:

0.111

Step-by-step explanation:

P{(X - 1.5)/3.2 < (-2.4-1.5)/3.2} = phi(-1.22)

Then you need to look into table called "Standard Normal Distribution Function"

In the table there are only positive values

For phi(1.22) = 0.8888

So for phi(-1.22) = 1 - 0.8888 = 0.111

At R programming:

pnorm(-2.4,1.5,sqrt(10.24)) = 0.1114695