A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30.
Find a 98% confidence interval for the mean price of all textbooks.

Step-by-step explanation:

We want to determine a 98% confidence interval for the mean price for textbooks.

Number of sample, n = 42

Mean, u = $114.5

Standard deviation, s = $12.3

For a confidence level of 98%, the corresponding z value is 2.33.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

114.5 ± 2.33 × 12.3/√42

= 114.5 ± 2.33 × 1.898

= 114.5 ± 4.42

The lower end of the confidence interval is 114.5 - 4.42 = 110.08

The upper end of the confidence interval is 114.5 + 4.42 = 118.92

Therefore, with 98% confidence interval, the mean price of all textbooks is between $110.08 and $118.92

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )