Question 1002065: Dear tutor,
I am having a problem with standard normal distribution table. I have been reading difference between positive and negative z values. but I am still having a hard time. For example, please see the question below.
Q.Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
I used calculator normalcdf (30, 10^99, 28, 5) = 0.3446
but the book calculated z score first 30-28/5 = 0.4 <-- I got this part!
From the normal table (this part is confusing)
Area to the right of 0.4 = 1- (0.5 + 0.15542) = 0.3446 <---- Not sure where this formula came from.If it is 0.4 why is not 0.6554 like on the table?
The required probability is 0.3446.
Answer is the same but trying to figure it out how to do it without using calculator....
b.Find the 95th percentile, and express it in a sentence.
InvNorm (0.95, 28, 5) = 36.2243
As explained in the book :
Area to the left = 0.95
From the normal table we need to find the area value closest to 0.95 - 0.5 = 0.45 and get the corresponding z-score. The closest match lower than 0.45 that we can get on the table is 0.44950 and the corresponding z-score value is 1.64.
- I do not understand this part at all.....what can't we just look at the table for 0.95?
Now we convert the z-score 1.64 to the x value using the formula (understand this formula)
z * sd = x - mean
x = z * sd + mean = 1.64 *5 + 28 = 36.2
thank you in advance!!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Dear tutor,
Q.Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
i believe you will find that the book may be giving you the half table rather than the full table.
there are 2 different types of tables most commonly used.
the full table shows you z-scores from - 3.5 or so to + 3.5 or so.
you read .95 directly from this table.
the half table shows you z-scores from 0 to + 3.5 or so.
this table is only showing you the right half of the distribution curve.
i have both tables and can show you how to do it from each.
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a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
m = mean
x = raw score
s = standard deviation
z = (x-m)/s
x = 30
m = 28
s = 5
z = (30-28)/5 = 2/5 = .4
using the full table:
look up a z-score of .4 and it will tell you that the area to the left of that z-score is .6554.
you want the area to the right of the z-score, so you take 1 - .6554 to get .3446.
using the half table:
look up the z-score of .4 and it will tell you that the area to the left of that z-score is .1554.
you want the area to the right of that z-score.
.5 - .1554 = .3446
same answer only you got it a different way.
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b.Find the 95th percentile, and express it in a sentence.
using the full table:
look up the area to the left of the z-score that is closest to .95
you will find .9495 and .9505
.9495 is a z-score of 1.64
.9505 is a z-score of 1.65
split the difference and you get a z-score of 1.645.
using the half table:
look up an area to the left of the z-score that is closest to .95 - .5 = .45
you have to subtract .5 because the area to the left of the z-score in the half table is only to the midpoint of the distribution curve because it is dealing only with the right side of the distribution curve.
the half table will tell you that the areas closest to .45 are .4495 and .4505.
those areas correspond to z-scores of 1.64 and 1.65.
split the difference and you get a z-score of 1.645.
same result as the full table only you got it a different way.
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your calculator assumes full table.
if you have to use the table, use the full table rather than the half table.
you can experiment with the half table and get your answer using that as well as the full table as well as the calculator.
the answers should be the same if you round to the same number of decimal places.
here's an example of the use of the half table compared to the full table.
how do you find the area to the left of -1.5?
with the full table, you look up -1.5 and it will tell you that the area to the left is equal to .0668
with the half table, you look up 1.5 and it will tell you that the area to the left is equal to .4332.
since the right half of the table is the mirror image of the left half of the table, you do not want the area to the left.
you want the area to the right.
using the half table, that would be .5 - .4332 = .0668
the area to the right of 1.5 is the same as the area to the left of -1.5
that's because the table is symmetric about the mean and the mean is always in the center of the normal distribution.
one more examle for good measure.
find the area between a z-score of -1 and 1.
using the full table:
find area to the left of 1 and area to the left of -1 and then subtract the smaller area from the larger area.
area to the left of z-score of 1 is .8413
area to the left of z-score of -1 is .1587
area between -1 and 1 is .8413 - .1587 = .6826
using the half table:
find area to the left of 1.
you will get area to the left of 1 is .3413
double that to get .6826
why double it?
left side is mirror image of right side.
area to the left of 1 is .3413
in the mirror image, this is the same as area to the right of -1.
area to the right of -1 is .3413
area between -1 and 1 is .6826.
you have to play with it for a while to become comfortable with it.
most times you will probably use a calculator.
most other times you will probably use a full table.
you would not normally use a half table unless that's all that is available.
here's some links to the tables i used plus a link to an online calculator you might find very interesting and useful.
FULL TABLE
HALF TABLE
CALCULATOR
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