Download MATH 215 Notes for Sections 6.2 and 6.3 Please read over both

MATH 215 Notes for Sections 6.2 and 6.3
Please read over both sections in the textbook yourself and read my notes below before trying the
homework.
Ch. 5 dealt with a discrete random variable (limited number of possible outcomes)
Ch.6 deals with a continuous random variable (unlimited number of possible outcomes)
Read over the information on p.251-2 regarding uniform distribution and density curves
The rest of the chapter deals with a normal distribution (bell shaped and approx. symmetric curve)
Section 6.2 starts off with the STANDARD normal distribution which has a mean of zero (which is the
center) and a standard deviation of 1 (scores are already standardized or z-scores) Recall z-scores tell us
how many standard deviations away from the mean a score is. Negative z-scores are below the mean
and positive z-scores are above the mean. Back in Ch. 3 we learned the empirical rule which
approximated the percentage of scores within 1, 2 and 3 standard deviations of the mean calling it the
68-95-99 rule for short since approx. 68% of scores fall within one standard deviation of the mean,
approx. 95% of scores fall within two standard deviations of the mean and approx. 99.7% of scores fall
within three standard deviations of the mean. In this section we are going to get more precise with
these percentages (probabilities) using Table A-2 in the book and/or technology. Please read the section
on pages 253-4 entitled “Finding Probabilities When Given Z Scores” – all five points along with table 6-1
too.
Note that Table A-2 in the very back of your book has two sides (one for negative and one for positive zscores). If you have a z-score and want to know the probability of scoring above or below that value,
you can look up the z-score on the table by first locating the ones and tenths value of the z-score in the
first column (vertical) of the table and then going over in that row (horizontal) to find the column that
has the correct hundredths value of the z-score. The number you find off the table is the area below
that z-score like shown in the diagram at the top of the table. The area under the density curve is the
same as the probability. For example to find the probability that a z-score would be less than 1.35
which would be written P(z < 1.35) you would go down to 1.3 in the first column of the table and go over
until you get to 0.05 and the table would give you .9115 (91.15% chance of occurring)and that is the
probability of getting a z-score less than 1.35 on a normal distribution.
How about if you want to find the P(z > 1.35)? That is the complement of above (since we don’t worry
about equality in continuous random variable cases) so you just need to subtract .9115 from 1 (since all
of the data falls on the chart) and you would get 0.0885 for the probability of getting a z-score greater
than 1.35.
How about if you wanted to find the P(-0.48 < z < 2.31)? That’s is you wanted to find the probability that
a z-score is between -0.48 and 2.31. You can look for the z-score of -0.48 on the negative z-score and
get .3156 but that is the probability that a z-score is below -0.48 and you don’t want that. When you
look up the z-score of 2.31 you get .9896 which is the probability that a z-score is below 2.31 and you
don’t want all of that either. So you take .9896 and subtract .3156 to get the area in between.
Now let’s go backwards. Suppose we want to find the 80th percentile (a score that separates the bottom
80% from the top 20%) for a standard normal distribution. So that would be a positive z-score since the
50th percentile is in the middle with a z-score of 0 and better than that would be a positive z-score and
worse than that would be a negative z-score. Now, we have the area (probability) below the z-score we
are trying to find. That area below is 0.8000 since it’s the 80th percentile. That is what we need to look
up in the body of the chart this time (where the four digit decimal numbers are) and find the closest
value to 0.8000 and follow it back to the appropriate z-score which in this case would be z = 0.84 since
.7995 is closer than .8023. Therefore, the 80th percentile corresponds to a z-score of 0.84.
How about if we want to find the score separating the bottom 4% from the top 96% (4th percentile)?
Now, we will look under the negative score table and look for the area (probability) below the score we
are looking for. Which would be 0.0400 (4%). So look it up in the body of the table and find the closest
value to 0.0400 which would be 0.0401 which corresponds to a z-score of -1.75. Therefore, a z-score of
-1.75 separates the bottom 4% from the rest of the data on a normal distribution.
(for more examples see pages 258-260 on finding z-scores from know areas (probabilities))
Section 6.3 deals with non-standard normal distribution which means we don’t already have z-scores or
(mean of 0 and standard deviation of 1) but we can standardize the data into z-scores using the z-score
formula we first learned in Ch. 3 (also on page 264 – formula 6-2). Take the raw score and subtract the
mean then divide that answer by the standard deviation.
So now we may be finding the probability of a randomly selected adult having a body temperature
above 101 degrees if we consider the mean body temperature to be 98.6 with a standard deviation of
0.82. In symbols we are finding: P(x > 101) but we need to find out what that means for the z-score.
So you take the 101 and subtract the mean and divide that difference by the standard deviation and get
a z-score of 2.93 (round z-scores to the nearest hundredth). So, P(x > 101) really means the same thing
as P(z > 2.93) in the standard normal distribution and we can use the table now. .9983 is the area
below 2.93 but we want the area above so subtract it from one to get 0.0017 which is the probability of
a randomly selected adult having a temperature above 101 degrees.
Here’s another example: Let’s say that the lengths of pregnancies are normally distributed with a mean
of 268 days and a standard deviation of 15 days. Find the probability that a randomly selected women
who has given birth had a premature birth (carried less than than 35 weeks = 245 days). In symbols we
are finding: P(x < 245) but we need to find how many standard deviations away from the mean the 245
is. So we take 245 – 268 and divide that answer by 15 to get a z-score of -1.53. So, P(x < 245) really
mean the same thing as P(z < -1.53) in the standard normal distribution. So, we look up a z-score of
-1.53 on the table and get 0.0630. So there is a 6.3% (approx.) that a randomly selected women had a
premature birth.
How about if we got the other direction again? Let’s use the same data from above regarding body
temperature (mean is 98.6 with a standard deviation of 0.82) but let’s find the temperature separating
the top 5% from the bottom 95% of people (95th percentile). We first look up the area below just like in
the last section to find the z-score. So we look up 0.9500 in the body of the chart and it is one of those
“special z-scores” since it falls exactly half way between the two values so the z-score is 1.645.
Remember that tells me how many standard deviations above or belwo the mean to go. So, we need to
go 1.645 standard deviations above the mean to get to the 95th percentile. So we take 1.645 x 0.82 and
add it to the mean of 98.6 to get 99.95.
Here’s another one. Let’s use the same data about the lengths of pregnancies (mean is 268 days with a
standard deviation of 15 days). This time let’s find what length separates the bottom 10% of women
from the top 90% of women for length of pregnancies. (find 10th percentile) So, we look up 0.1000 in
the body of the negative z-score table and find the closest value which is 0.1003 which corresponds to a
z-score of -1.28. So, that tells us we need to go below the mean by 1.28 standard deviation to get to the
10th percentile. So we take 1.28 x 15 and subtract it from the mean of 268 to get 248.8 days which is
approx. 35.5 weeks.
I hope these notes help but I’d much rather be there in person explaining it. See you Thursday!! Please
let me know if you have any questions.