Question

A project is found to have expected time T = 35.33 days and variance V = 3.22.

Answer 1

Answer:

a) $z = \frac{40-35.33}{1.794}= 2.60$

b) $z = \frac{40-35.33}{1.794}= 2.60$

c) $z = \frac{30-35.33}{1.794}= -2.97$

Step-by-step explanation:

For this case we know that the mean for the random variable of interest is $\mu = 35.33$ and the variance $\sigma^2 = 3.22$ so then the deviation would be $\sigma = \sqrt{3.22}= 1.794$

The z score is given by thsi formula:

$z = \frac{X -\mu}{\sigma}$

Part a

We want this probability:

$P(X>40)$

And if we find the z score we got:

$z = \frac{40-35.33}{1.794}= 2.60$

And we can find this probability: $P(Z>2.60)$

Part b

We want this probability:

$P(X$

And if we find the z score we got:

$z = \frac{40-35.33}{1.794}= 2.60$

And we can find this probability: $P(Z$

Part c

We want this probability:

$P(X$

And if we find the z score we got:

$z = \frac{30-35.33}{1.794}= -2.97$

And we can find this probability: $P(Z$