Answer:
C.
Step-by-step explanation:
The answer to your question would be C. No, the data values in each class could take on any value between the class limits, inclusive.
I hope it helps! Have a great day!
Muffin~
Answer:
A.) The manager calculated mean and not standard deviation
B.) Standard deviation should be calculated with the use of formula.
Step-by-step explanation:
A.) The assumption that his customers do spend more than the national average is wrong because the standard deviation is not calculated and mean cannot be used as a substitute for standard deviation.
B.) Standard deviation tells us how value obtain from group measurement deviates from the average value or expected value of each item of observations.
When standard deviation is low, this show that the value is very close to the expected value or the average value.
If the national average value = $150 and average customer spends $160.
Deviation = 160 - 150 = 10
This is lower than $30
Or better still, the manager should make use of standard deviation formula. Which is the square root of variance.
Answer:
37x, 12
Step-by-step explanation:
Using the normal distribution, there is a 0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean
and standard deviation
, as long as
and
.
The proportion estimate and the sample size are given as follows:
p = 0.45, n = 437.
Hence the mean and the standard error are:
The probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3% is <u>2 multiplied by the p-value of Z when X = 0.45 - 0.03 = 0.42</u>.
Hence:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem:
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
Z = (0.42 - 0.45)/0.0238
Z = -1.26
Z = -1.26 has a p-value of 0.1038.
2 x 0.1038 = 0.2076.
0.2076 = 20.76% probability that the proportion of persons with a college degree will differ from the population proportion by greater than 3%.
More can be learned about the normal distribution at brainly.com/question/28159597
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