Answer:
Confidence Interval in 95% confidence level for the quality rating is (6.06,7.46)
Step-by-step explanation:
Confidence Interval can be calculated using the formula M±ME where
- M is the mean of the sample
- ME is the margin of error in a given confidence level
Using the sample obtained from 50 business travelers we get
- Mean of the sample is 6.76
- standard deviation of the sample is 2.526
Margin of error (ME) around the mean using the formula
ME=
where
- z is the corresponding statistic in 95% confidence level (1.96)
- s is the standard deviation of the sample (2.526)
- N is the sample size (50)
Using the numbers in the formula we get:
ME=
≈ 0.70
Then the confidence interval becomes 6.76±0.70
The ACT score was better since the z-score of the ACT is greater than the z-score of the SAT. Then the correct option is B.
<h3>What is a normal distribution?</h3>
It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.
The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.
The z-score is given as

In 1985, the average ACT score was 18 with a standard deviation of 6.
Also in 1985, the average SAT score was 500 with a standard deviation of 100.
Dr. Robertson took both tests in 1985 and scored a 26 on the ACT and a 620 on the SAT.
Then the z-score of the ACT will be

Then the z-score of the SAT will be

Then the ACT score was better since the z-score of the ACT is greater than the z-score of the SAT.
More about the normal distribution link is given below.
brainly.com/question/12421652
#SPJ1
Option B is the answer and that because the photo below
First we use product rule
y=x^2lnx
dy/dx = x^2 d/dx (lnx) + lnx d/dx (x^2)
dy/dx = x^2 (1/x) + lnx (2x)
dy/dx = x + 2xlnx
now taking second derivative:
d2y/dx2 = 1 + 2[x (1/x) + lnx (1)]
d2y/dx2 = 1 + 2[1+lnx]
1+2+2lnx
3+2lnx is the answer