Given the three integers are , we arrive at
We can merge the fractions on the left side:
All computers are on sale for 10% off the original price. If x is the original price of the computer, then the function that rep
2 answers:
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Response:
Detailed explanation:
Greetings!
You have the variable
X: Area eligible for painting with a can of spray paint (feet²)
This variable is normally distributed with a mean of μ= 25 feet² and a standard deviation of δ= 3 feet²
As this variable has a normal distribution, it needs to be converted into the standard normal form to utilize tabulated cumulative probabilities.
a.
P(X>27)
The first step involves standardizing the X value using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Having determined the Z value, you can find it in the table, but since the table includes probabilities for , the following conversion must be applied:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
A sample of 20 cans was taken, and you need to ascertain the probability of averaging a coverage area of 540 feet².
The sample mean maintains the same distribution as its source variable, but its variance is influenced by sample size, thus it is normally distributed with parameters:
X[bar]~N(μ;δ²/n)
To cover 540 feet² with 20 cans, the average coverage must be approximately 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999d.No, if the distribution is not normal and skewed, the normal distribution should not be applied for calculating probabilities. While the central limit theorem might approximate the sampling distribution to normal when the sample size is 30 or larger, that isn’t applicable here.
I trust this information is helpful!
Answer:
Step-by-step explanation:
It is known that the mean and standard deviation of the sampling distribution of the sample proportion() are represented as follows:-
, where p= Population proportion and n = sample size.
Let p denote the proportion of blue chips.
According to the information provided, we have
p= 0.275
n= 5
Thus, the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 will be:
Therefore, you will have the mean and standard deviation for the sample proportion of blue chips for samples of size 5: