Week 7 Quiz Help Things to remember for this week’s quiz. There’s some good info here, pay close attention. With the standard normal distribution Remember that the total area under the curve is equal to 1 or 100% (half on each side). The mean of the standard normal is 0 and the standard deviation and variance are 1. For any normal distribution regardless of the mean and standard deviation the area will always be 1 or 100%. We see the normal distribution all around us as noted in one of our discussion topics this week. Also remember that if you ever told that you have a normal distribution and then asked what the probability x = ?, it is zero. We can give the probability that it is less than a value, greater than a value, or between two values – but not that it is exactly one value because it is a “continuous distribution”. (For example given that mu = 9, sigma = 2.1, what is the probability that x = 7? It’s 0 because with a continuous distribution we are “slicing jello” as I like to say. On the Central Limit Theorem, remember that it states that given a distribution with a mean μ and variance σ², the
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1. Week 7 Quiz Help Things to remember for this weeks quiz.
Theres some good info here, pay close attention. With the standard
normal distribution Remember that the total area under the curve is
equal to 1 or 100% (half on each side). The mean of the standard
normal is 0 and the standard deviation and variance are 1. For any
normal distribution regardless of the mean and standard deviation
the area will always be 1 or 100%. We see the normal distribution
all around us as noted in one of our discussion topics this week.
Also remember that if you ever told that you have a normal
distribution and then asked what the probability x = ?, it is zero.
We can give the probability that it is less than a value, greater
than a value, or between two values but not that it is exactly one
value because it is a continuous distribution. (For example given
that mu = 9, sigma = 2.1, what is the probability that x = 7? Its 0
because with a continuous distribution we are slicing jello as I
like to say. On the Central Limit Theorem, remember that it states
that given a distribution with a mean and variance , the sampling
distribution of the mean approaches a normal distribution with a
mean () and a variance /N as N, the sample size, increases. The
amazing and counter-intuitive thing about the central limit theorem
is that no matter what the shape of the original distribution, the
sampling distribution of the mean approaches a normal distribution.
Also another important fact is that any sample size is big enough
when we know the population is normal. Using the Central Limit
Theorem answer the following question.
2. Assuming you have a normal distribution for your population
and you take 64 samples of size 25 each. Calculate the standard
deviation of the sample means if the populations variance is 16.
Since the population is normally distributed with a variance of 16,
then the sample means have a variance equal to 16/25 according to
the Central Limit Theorem. Hence their standard deviation will be
SQRT(16/25) = 4/5 = .800 Be able to use the normal distribution to
solve problems. Examples Bob scored a 190 on his entrance exam,
where the average was 165 and the standard deviation was 12. Where
does he stand in relation to the rest of his class? He scored in
the top 2 %, see excel attachment. In a normal distribution with mu
= 35 and sigma = 6 what is the z score for a value of 41? Z= +1 see
excel attachment In a normal distribution with mu = 35 and sigma =
6 what number corresponds to z = -2? 23, see excel attachment We
have an area of .4840. What z-score corresponds to this area?
3. Using the Standard Normal Table in your book, or the one on
the attached excel you can see it is a z score of -0.04 Find P(80
< x < 86) when mu = 82 and sigma = 4. Write your steps in
probability notation. I did this in excel (see attachment fourth
tab), but I still need to show my work: The z-score corresponding
to x = 86 is z = (86-82)/4 = 4/4 = 1.0. The area corresponding to z
= 1.0 is .8413 The z-score corresponding to x = 80 is (80 - 82)/ 4
= -2/4 = -0.5. The area corresponding to z = -0.5 is .3086. Thus,
P(80 < x < 86) = P(-0.5 < z < 1.0) = P(z < 1.0) -
P(z < -0.5) = 0.8413 0.3086 = 0.5327 (my excel calculated it to
be 0.5328 so I feel good about it) My excel file will be in a
separate post.
4. Interpret a 90% confidence interval of (63.3, 83.4). You
should note that there here is a 90% probability that the interval
(63.3 to 83.4) contains , the true population mean. What is the
critical value that corresponds to a confidence level of 96% 100 96
= 4 Divide 4 by 2 (tails) and get 2 Add 2 to the original 96% and
get 98% and find the critical value (z-score that corresponds to .
9800 which is 2.05 (closest to it) Compute the population mean
margin of error for a 90% confidence interval when sigma is 7 and
the sample size is 81. E = z * sigma / sqrt(n) = 1.645 * 7 /
sqrt(81) = 1.279 (remember +/-)
5. A Military entrance exam has a mean of 120 and a standard
deviation of 9. We want to be 95% certain that we are within 6
points of the true mean. Determine the sample size. n = ( z * sigma
/ error ) ^ 2 = (1.96*9/6)^2 = 2.94^2 = 8.6436. Round up to 9. A
researcher wants to get an estimate of the true mean performance
measure of its product. It randomly samples 180 of its machines.
The mean performance measure was 900 with a standard deviation of
60. Find a 95% confidence interval for the true mean performance
measure of the machines. The population standard deviation is
unknown and the sample size is 180. Thus, since the sample size is
greater than 30, this confidence interval will use a z-value. For a
95%
6. confidence interval, the z-value = 1.96. Sample mean = 900
and sample standard deviation = 60. Population mean = 900 +/- 1.96
* 60/sqrt(180) = 900 +/- 8.765. 891.235 and 908.765 A researcher
wants to get an estimate of the true mean performance measure of
its product. The researcher needs to be within 15 of the true mean.
The researcher estimates the true population standard deviation is
around 30. If the confidence level is 95%, find the required sample
size in order to meet the desired accuracy. For a 95% confidence
level, the z-value = 1.96. The formula for sample size is n = (
z-value * standard deviation / error ) ^ 2 = ( 1.96 * 30/ 15) ^ 2 =
( 3.92 ) ^ 2 = 15.3664. Thus, the researcher must sample at least
16 to obtain the desired accuracy.
7. A researcher wants to estimate the mean cost to develop a
product. The researcher tests 18 cases and finds the mean cost to
be $3000 with a standard deviation of $400. Find a 95% confidence
interval for the true mean cost to develop this product. The
population standard deviation is unknown and the sample size is 18.
Thus, since the population standard deviation is unknown AND the
sample is less than 30, we must use the t- value for this
confidence interval. For a 95% confidence interval and degrees of
freedom = 17 (from 18-1), the t-value = 2.110. Sample mean = 3000
and sample standard deviation = 400. Population mean = 3000 +/-
2.110 * 400/sqrt(18) = 3000 +/- 198.93. So our bounds are 2801.07
and 3198.93
8. A researcher wants to estimate what proportion of failures
that are due to poor workmanship. The researcher randomly samples
50 failures and finds 18 are due to poor workmanship. Using a 95%
confidence interval, estimate the true proportion of poor
workmanship for all failures. For a 95% confidence level, the
z-value is 1.96. The sample proportion = 18/50 = 0.36, thus p hat =
0.36 and 1-p hat = 0.64 . The sample size = 50. The population
proportion is between 0.36 +/- 1.96 * sqrt ( 0.36 * 0.64 / 50 ) =
0.36 +/- .13 So the bounds are .23 and .49 Excel Spreadsheets
Separately
9. Normal Distribution Me a n S tde v 165 12 P (Xx) x1 P (x 1
x) S ymme tric Inte rva ls x1 P (x 1