Dr. Tease is a professor of English. Recently she counted the number of misspelled words in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a population standard deviation of 2.44 words per essay. For her 10 a. m. section of 40 students, the mean number of misspelled words was 6.05. Construct a 95% confidence interval for the mean number of misspelled words in the population of student essays.

The 95% confidence interval for the mean number of misspelled words in the population of student essays is between 5.29 and 6.81.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 6.05 - 0.76 = 5.29

The upper end of the interval is the sample mean added to M. So it is 6.05 + 0.76 = 6.81

The 95% confidence interval for the mean number of misspelled words in the population of student essays is between 5.29 and 6.81.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

answer: n word

step-by-step explanation:

what type of people are in the hood?

what steals your tv in the dark and you cant see them?