Random Variable Z

[willster459]

I am trying to do these problems and I am so lost so if anyone could give me a little guidance it would be so amazing!!

Find the following probabilities for the standard normal random variable z:

P(z>1.46)

P(z< -1.56)

P(.67 ? z ? 2.41)

P(-1.96?z< -.33)

P(z?0)

P(-2.33<z<1.50)

P(z ? -2.33)

P(z < 2.33)

Find a value of the standard normal random variable z, call it z (subscript) 0, such that

P(z ? z subscript 0)=.0401

P(-z subscript 0 ? z ? z subscript 0) =.95

P(-z subscript 0 ? z ? z subscript 0)=.90

P(-z subscript 0 ? z ? z subscript 0)=.8740

P(-z subscript 0 ? z ? 0)= .2967

P(-2 < z < z subscript 0) = .9710

P(z ? z subscript 0)= .5

P(z ? subscript 0)=.0057

Thanks in advance to anyone who can help!!

 

[galactus]

Find the following probabilities for the standard normal random variable z:

Here are a few to get you started. Once you catch on this is very easy.

P(z>1.46)

Look them up in the z table.

Since the z table comes from negative infinity and heads right, look up the value in the z-table corresponding to 1.46 and subtract from 1.

1-.9278=.0722

When they ask for a greater than, look up the value and subtract from 1.

P(.67 ? z ? 2.41)

P(.67)=.7486

P(2.41)=.9920

Subtract the two values because we want the area between them.

.992-.7486=.2434

P(z?0)

0 is in the center of the curve. Since the area under the whole curve is 1, then we have half that. No table necessary.

Here are the graphs for the first two problems to help visualize.