Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:
And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Answer:
35
Step-by-step explanation:
You have 17 either side to make 18 the middle e.g if 2 was the middle then you have 1 either side.
Answer:
The <em>p-</em>value of the test is 0.106.
The null hypothesis will be accepted at 5% level of significance.
Step-by-step explanation:
The hypothesis test is left-tailed.
The test statistic value is: <em>z</em> = -1.25.
The significance level of the test is: <em>α</em> = 0.05.
The <em>p</em>-value of a left-tailed hypothesis test is:
The <em>p-</em>value of the test is 0.106.
**Use the <em>z-</em>table for the probability.
<u>Decision rule:</u>
If the <em>p</em>-value is less than the significance level the null hypothesis will be rejected and if it is more than the significance level the null hypothesis will be accepted.
The <em>p</em>-value = 0.106 > <em>α</em> = 0.05.
The null hypothesis will be accepted at 5% level of significance.
THE CORRECT ANSWER WAS
$160.26
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