Statistics...z-score for given probability in tails?

For a normal distribution,





a. Find the z-score for which a total probability of 0.02 falls more than z standard deviations (in either direction) from the mean.





b. For this z, explain why the probability more than z standard deviations above the mean equals 0.01.





c. Explain why µ + 2.33σ is the 99th percentile.





*Note:


The z-score for a value x of a random variable is the number of standard deviations that x falls from the mean µ. It is calculated as





z = (x- µ) / (σ)

Statistics...z-score for given probability in tails?
a)


P(z %26gt; Z) = .01 implies Z = 2.326





b)


The area under the normal curve beyond 2.326 is equal to .01. This Z satisfies part a because the area beyond -2.326 is also .01. Summing these areas in the tails together is .02. We consider both tails because the question stated "in either direction."





c)


µ + 2.33σ is the result of solving the z-score formula for x. So we could write: µ + 2.33σ = x


µ and σ are constants that determine the center and spread of a particular normal curve. The z-value of 2.33 corresponds to a value of x on this normal curve such that 99% of all possible values of X will be less than said x value. Symbolically: P(X %26lt; x) = .99
Reply:a)


P( -z %26lt; Z %26lt; z ) = 0.02


z = 2.326348





b)


you have a two tail situation in part a, so you have 1% on either tail.





c)





because P(Z %26lt; 2.33) = 0.99