5h−2(−8h+9), when h=8
1 answer:
You might be interested in
Answer: a.) $50188 to $57812
Step-by-step explanation: <u>Confidence</u> <u>Interval</u> (CI) is an interval of values in which we are confident the true mean is in.
The interval is calculated as
x ±
a. For a 95% CI, z-value is 1.96.
Solving:
54,000 ±
54,000 ±
54,000 ± 1.96*1732.102
54,000 ± 3395
This means the interval is
50605 < μ < 57395
<u>With a 95% confidence interval, the mean starting salary of college graduates is between 50605 and 57395 or </u><u>from 50188 to 57812$.</u>
<u />
b. The mean starting salary for college students in 2017 is $50,516, which is in the confidence interval. Therefore, since we 95% sure the real mean is between 50188 and 57812, there was no significant change since 2017.
I got D.
There's a few ways to solve it; I prefer using tables, but there are functions on a TI-84 that'll do it for you too. The logic here is, you have a standard normal distribution which means right away, the mean is 0 and the standard deviation is 1. This means you can use a Z table that helps you calculate the area beneath a normal curve for a range of values. Here, your two Z scores are -1.21 and .84. You might notice that this table doesn't account for negative values, but the cool thing about a normal distribution is that we can assume symmetry, so you can just look for 1.21 and call it good. The actual calculation here is:
1 - Z-score of 1.21 - Z-score .84 ... use the table or calculator
1 - .1131 - .2005 = .6864
Because this table calculates areas to the RIGHT of the mean, you have to play around with it a little to get the bit in the middle that your graph asks for. You subtract from 1 to make sure you're getting the area in the middle and not the area of the tails in this problem.
