The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi
1 answer:
Answer:
a) Mean = 1030; Standard deviation = 12.38.
b) The county result is unusually high.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.
This first question is a binomial propability distribution.
We have a sample of 1210 Amricans, so .
The mean of the sample is 1030.
The probability of a success is .
The standard deviation of the sample is
(b) Is that county result of 1030 unusually high, or low, or neither?
The first step is find the zscore when .
Then we find the pvalue of this zscore.
If this pvalue is bigger than 0.95, the county result is unusually high.
If this pvalue is smaller than 0.05, the county result is unusually low.
Otherwise, it is neither.
The national mean is 82%. So,
has a pvalue of 0.9989.This means that the county result is unusually high.
You might be interested in
Answer:
a
b
Step-by-step explanation:
From the question we are told that
The mean is
The standard deviation is
Generally 2 year is equal to 24 months
Generally the percentage of total production will the company expect to replace is mathematically represented as
Generally
Generally from the z-table
So
Converting to percentage
=>
Generally the duration that should be the guarantee period if Accrotime does not want to make refunds on more than 6% is mathematically evaluated as
=>
From the normal distribution table the z-score for 0.06 at the lower tail is
So
=>
Answer:
Step-by-step explanation:
Hello!
You have the variable
X: Area that can be painted with a can of spray paint (feet²)
The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²
since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.
a.
P(X>27)
First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of , you have to do the following conervertion:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².
The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:
X[bar]~N(μ;δ²/n)
So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)
To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999
d.
No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.
I hope it helps!