Answer:
The random variable W is Normal, with a mean of 48 ounces and a standard deviation of four ounces.
Explanation:
By their properties, mean of μ(4x) = 4 * μ(x) and a standard deviation of σ(4x) = 4 * σ(x).
So the distribution of weight of four loaves of bread, W, still follows a Normal Distribution with a mean of 4*12 = 48 ounces and a standard deviation of 4*1 = 4 ounces.
The following statements describes the random variable W is Normal, with a mean of 48 ounces and a standard deviation of four ounces.
Given the information that:
distribution with a mean of μ = 12 ouncesstandard deviation of σ = 1 ounce.pick four loaves at randomBy their properties, we have:
[tex]\mu(4x) = 4 * \mu(x) \\\sigma(4x) = 4 * \sigma(x).[/tex]
So the distribution of weight of four loaves of bread, W, still follows a Normal Distribution with a mean:
[tex]4*12 = 48 \\4*1 = 4[/tex]
See more about distribution at brainly.com/question/14926605
On a certain portion of an experiment, a statistical test result yielded a p-value of 0.15. What can you conclude?
If the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 15% of the time, so the test is not statistically significant.
2(0.15) = 0.30 < 0.5; the test is not statistically significant.
0.15 > 0.05; the test is statistically significant.
If the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 85% of the time, so the test is not statistically significant.
p = 1 - 0.15 = 0.85 > 0.05; the test is statistically significant.
Answer:
15%
Explanation:
Answer:
If the null hypothesis is true, one could expect to get a test statistic at least as extreme as that observed 15% of the time, so the test is not statistically significant.
Explanation:
Got this right on the segment exam. Good luck, my FLVS fellows!
how
do i put a picture in my question
Answer: click on the paperclip icon and your files should pop up :)
Explanation: