Chapter 2.2, Problem 51E

(a)

To determine

To find: the valuez from the standard Normal distribution

Answer to Problem 51E

The value of z corresponding to 10^thpercentile is -1.28.

Explanation of Solution

Given:

The 10^th percentile.

Calculation:

To find the z-score that corresponds to 10' percentile find the z-score that corresponds toan area 0.10 by locating 0.10 in the Standard Normal probabilities table. The values at the beginning of the corresponding row and at the top of the corresponding column give the z-score.

  

From the Standard Normal probabilities table, the area closest to 0.10 in the table is 0.1003(z =-1.28).

So, the z-score (approximately) that corresponds to an area of 0.10 is -1.28.

The following graph illustrates 10^th percentile along with the corresponding z value as follows:

  

Hence, the value of z corresponding to 10^th percentile is -1.28.

Conclusion:

Hence, the value of z corresponding to 10^th percentile is -1.28.

(b)

To determine

To find: the valuez from the standard Normal distribution

Answer to Problem 51E

The value of z such that 34% observations are above z is 0.41.

Explanation of Solution

Given:

34% of all observations are greater than z.

Calculation:

We are supposed to find the value of z such that 34% of all observations are greater thanz .

If 34% of observations are above z values, it means that there exists 66% of observations are below z.

  

From the Standard Normal probabilities table, the area closest to 0.66 in the table is 0.6591

( z =0.41).

So, the z-score (approximately) that corresponds to an area of 0.66 is 0.41.

The result is shown in the below graph.

  

Hence, the value of z such that 34% observations are above z is 0.41.

Conclusion:

Hence, the value of z such that 34% observations are above z is 0.41.