Download A sample of observations is drawn from a normal population with

Mth 243 Suggestions for CLT Assignment in WW
1. To get answers that WebWork likes, caculate the z values and round to two decimal
places. Then, look this z-value up in the table or use your calculator.
2. To use your calculator to convert z values into probabilities, the command is:
normalcdf (lower bound, upper bound, 0, 1)
The 0,1 represents a standard normal distribution with mean 0 and std
deviation 1. It is optional!
Example: P(z < 1.28) would be entered normalcdf(-9999, 1.28,1,0) or simply
normalcdf(-9999, 1.28)
Example: Given µ = 55 , σ = 8 and population normally distributed… Skip zvalue calculation. The P(x<67) = normalcdf(-9999, 67, 55, 8)
Warning: This last method may be compatible with WW but not always.
3. You can still use the table but you may need to extrapolate between probabilities
listed in the table.
Problem 5
(Read the suggestions above.)
A sample of
with
observations is drawn from a normal population
and
. Find the following probabilities:
????
Well, the mean of the sample means is the same as the mean of the
population. Mathematically that looks like
µx = µx . The sampling error is not
the same as the population standard deviation, σ . The standard error is σ / n .
So, the sampling distribution can be described as:
(
)
N 1000, 200 / 16 or as N (1000, 50 )
Pr ob( x > 1050) = Pr ob( z >
1050 − 1000
) = Pr ob( z > 1)
50
In class a few weeks back we learned that this probability was about .16 . But this is
approximate and not good enough for WebWork. So, use your calculator… Enter
normalcdf(1, 9999) to get the answer: 0.1587 Can you replicate my answer?! WebWork
likes this answer!
Or, more directly: normalcdf(1050, 99999999, 1000, 50) = 0.1587 Why did I use 50
rather than σ = 200 ? The 50 is the sampling error.
normalcdf(-99999, -.8) or normalcdf(-999999, 960, 1000,50)
Problem 6
Given a normal distribution whose mean is 50 and whose standard deviation is 5.
Find the probability that a random sample of 4 has a mean between 49 and
52.
Mathematically: Pr ob(49 < x < 52) Enter: normalcdf(49, 52, 50, 2.5)
WW took my answer: .4436 no problem.
Find the probability that a random sample of 25 has a mean between 49 and
52.
Mathematically: Pr ob(49 < x < 52) Enter: normalcdf(49, 52, 50, 1)
WW took my answer: .8186 no problem.
Problem 9
p and pˆ !!!
Remember that p̂ has an approximately normal distribution only if the following
This is exciting -
conditions are met: np and n(1-p) are both greater than 5.
In this problem p = 0.32 and n = 100. So clearly .32(100) and .68(100) are both above 5!
Therefore, the sampling distribution can be described mathematically as:
Notice that p is in the place of the mean, and that the standard error is that square root
expression! I calculated the SE = 0.466 but WebWork wanted me to be more accurate,
SE = 0.04665. OK, I can do that.
The sampling distribution becomes:
N (0.32,.04665)
Finally, find the probability that more than 34 percent of women sampled will say they
believe in aliens.
That is:
Pr ob( pˆ > .34) = normalcdf (.34,99999,0.32, 0.04665)
My answer 0.3341 was rejected. But, if I calculated a z-value and rounded the z-value to
two decimal places, it worked! (Sorry.) i.e. normalcdf(0.43, 99999) equals .3336. WW
liked this answer.
Problem 10
It is estimated that approximately 55% of college students graduate in 5 years
or less. This figure is affected by the fact that more students are attending
college on a part-time basis. If 500 students on a large campus are selected at
random, what is the probability that between 52% and 62% of them will
graduate in 5 years or less?

 .52 − .55
.62 − .55
Pr ob(.52 < pˆ < .62) = Pr ob 
<z<
 .55(.45)
.55(.45)

500
500



=



.07 
.07 
 −.03
 −.03
<z<
<z<
Pr ob 
 = Pr ob 
=
0.02225 
0.02225 
 0.02225
 0.02225
Pr ob( −1.35 < z < 3.15) = .9107 An answer WW likes.
Remember, WW likes you to roundoff only when you get to your z-value. That way you
can use the calculator or your table to find probabilities.
Problem 12
Since cell phone plans have been introduced, the data shows that the average
number of minutes used per month on a "family" plan is 888 minutes, with a
standard deviation of 260 minutes. In a random sample of 50 "family" plans,
what is the probability that more than 1010 would be used?
(
The sampling distribution is: N 888, 260 / 50
)
We want the Pr ob( x > 1010) = Pr ob( z > 3.32) = .0005
I rounded my calculation of .00045 to .0005. WW wanted my final answer to be rounded
to the “ten thousandth”.
Problem 13
A manufacturer of household appliances claims that only 2% of all its products
require a service call in the first year. A consumer protection group wants to
verify the claim and to do this, it surveys 350 households where one of these
manufacturer's appliances had recently been purchased.
this is saying that the mean of the p-hats is the same as p , .02
this is saying that the standard error of p-hat is
p(1 − p )
= 0.00748
n
I sure hope this helps. Read your text book as well! Good luck - Rob