THis is homework can you help?
Suppose that the heights of adult women in the United States are normally distributed with a mean of 64.5 inches and a standard deviation of 2.4 inches. Jennifer is taller than 80% of the population of U.S. women. How tall (in inches) is Jennifer? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.
There are two different kinds of normal tables that are used,
and some people don't use tables at all, but either the TI-84
calculator or the computer program Excel.
Using tables:
If you have the kind of table that has both negative and positive z
values in it, then you'll find the values .7995 and .8023, because 80%
or .8000 is between those two numbers. That is, the z-value entry which
is closest to having 80% of the area to the left of it. It's a matter of
looking through the body of the table until you find the entries which
.8 is between: Those listings correspond to z-values 0.84 and 0.85.
Since .7995 is closer to .8000 than .8023, we choose z-value 0.84.
If you have the kind of table that has only positive z-values, then you
subtract .8 - .5, getting .3. then you'll find the values .2995 and .3023,
because .3 (or .3000) is between those two numbers. That is, the z-value
entry which is closest to having 30% of the area to the right of the
middle. the left of it. It's a matter of looking through the body of the
table until you find the listings correspond to z-values 0.84 and 0.85.
Since .2995 is closer to .3000 than .3023, we choose z-value 0.84.
From either table, we get z = 0.84
Then we use the formula
x = m + sz
x = 64.5 + (2.4)(0.84) = 66.516
If you use a TI-83 or 84 calculator,
Press
2nd
VARS
3
Then type
.8
, the comma key
64.5
, the comma key again
2.4
) the close parenthesis key
(The screen should now read invNorm(.8,64.5,2.4)
press
ENTER
Read 66.51989096
Edwin
