(as long as the population is large compared to the sample (at least 10 to 20 times sample)
The mean of the x-bars = the mean of the population
The standard deviation of the x-bars =
the s. d. of the population divided by the square root of n.
| Hand in Also, Complete your HW from Day 30 if
you need to. Try to do 11.39 and 38 ("backward" problems) (I did 38 in
class, so it should be possible) , but don't lose sleep over them. Be
sure you can do the "forward" types (36, 37).
These problems use the Central Limit theorem (p. 281) p. 185, 11.10 What does the CLTh say? p. 286 , 11. 12 SAT scores, n = 1 and 70 p. 286, 11.13, insurance (Hint: find P(Xbar> $275)) p. 298, 11.41 auto accidents p. 298, 11.42 airplane overloads (Hint: to do the problem you have to assume all the seats are taken. Maybe not a reasonable assumption, but if there are empty seats, there's likely not a problem with overweight.) |
Read, to discuss |
Optional |
We talked about the above Friday,
up to Central Lim. th. HW questions Day 30?
B) "Normal" body temperature 98.6 deg. on average. (Assume this is
true.)
Assume normal distribution, & s.d.among many
people is 0.6.
Probability that one
(random) healthy individual's normal temperature is above 98.8?
Probability that the mean of a sample
of n is above 98.8?
All of these are P(Xbar>98.8) for
different sample sizes n. (Normal
table A)
| Sample size n |
s.d. of Xbars = (pop.s.d.)/sqrt(n) |
z = (raw-mean)/s.d. |
P(Xbar>98.8)= P(Z>z) |
| 1 |
.6/1 = .6 |
(98.8-98.6)/.6 = .2/.6 = .33 |
P(Z>.33) = .3707 |
| 4 |
.6/2 = .3 |
(98.8-98.6)/.3 = .2/.3 = .67 | P(Z>.67) = .2514 |
| 36 |
.6/6 = .1 |
(98.8-98.6)/.1 = .2/.1 = 2 | P(Z>2) = .0228 |
| 100 |
.6/10 = .06 |
(98.8-98.6)/.06 = .2/.06 = 3.33 | P(Z>3.33) = .0004 |
Excel pictures for these Xbar distributions. Normal X vs Xbar
Today, the Central Limit Theorem Day
29, details
Central Limit Theorem...
How large is "large"? How approximate is "approximate"?
If the population was close to normal,
n doesn't need to be very large.
If the population is not badly skewed
or bimodal, n=25 already gives a pretty good approximation to normal.
Got about here Monday. You can do all the
HW above. Next time I'll show more examples, to help you see how/that the
distribution of Xbars ("sampling distribution of the mean") is approx.
normal when n is large.
Sum of 3 dice (divide bottom scale by 3 to get average
of 3 dice)
Pictures, n=1, 2, 4, 25 CLTpics.pdf
What if the population is not 10 to 20 times the sample size? The
real s.d. of the x-bars will be narrower than the rule above. You
may not "get to" normal as a shape. Sample of 4 grades from a population
of 10: This and one other semester (big file,
loads slowly off campus) last year All possible samples. ...
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