The quotient of a number and 12 is no more than 6 \
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
It was noted that a music streaming platform modified its format to highlight previously unreleased tracks from emerging artists. The site manager is now aiming to assess whether the daily unique listener count has changed.
The hypothesis is set
(A two-tailed test for mean difference)
The test statistic is calculated, and the p-value turns out to be 0.0743
Assuming a significance level of 5%, we observe that p-value = 0.0743>0.05Thus, we accept the null hypothesis.i.e. there is no statistically significant alteration in the average number of daily unique listeners
The p-value serves as an indicator of the extremity of the observed data. If p is lower than alpha, we thus reject H0; otherwise, we accept it.
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!