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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.
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Answer:
M=2,179.29
The average exceeds the estimate
Her expenses would need to be $1,625.00
Step-by-step explanation:
1.) What is the mean of her friends' personal expenses?
The mean is defined as the average expense value, calculated as follows:
2.) Is the average higher or lower than the estimate?
Given the estimate of $2,110, the average is indeed higher than this estimate.
3.)What amount of personal expenses would Ashley need to maintain a combined average of $2,110 with her friends' amounts for that school year?
Her needed amount of expenses would be $1,625.00.
Answer:
a) Null and alternative hypothesis
b) Point estimate d = -$78
c) Test statistic t = -2.438
P-value = 0.0113
Reject H0. This indicates that the average automobile premium in Pennsylvania is lower than in the nation.
Step-by-step breakdown:
This is a statistical test for the average population mean.
The hypothesis posits that car insurance in Pennsylvania is notably less expensive compared to the national average.
Accordingly, the null and alternative hypotheses are:
The significance level is set at 0.05.
The sample size is n=25.
The sample mean equates to M=1425.
A point estimate of the difference between the Pennsylvania mean premium and the national average can be computed using the sample mean:
Given that the standard deviation of the population is unknown, we approximate it using the sample standard deviation, which is s=160.
The estimated standard error of the mean is determined with the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size stand at:
This constitutes a left-tailed test, with 24 degrees of freedom and t=-2.438, rendering the P-value as (per a t-table):
As the P-value (0.0113) falls below the significance level (0.05), the results prove significant.
Thus, the null hypothesis obtains dismissal.
At a 0.05 significance level, there's sufficient evidence to assert that car insurance in Pennsylvania costs notably less than the national average.