I just need help on #8.
2 answers:
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Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
<h3>Normal Probability Distribution</h3>In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation
.
In this problem:
- The mean is of 660, hence
.
- The standard deviation is of 90, hence
.
- A sample of 100 is taken, hence
.
The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:
By the Central Limit Theorem
has a p-value of 0.8665.
1 - 0.8665 = 0.1335.
0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.
To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213
Answer:
Step-by-step explanation:
Assuming conditions are met, the formula for a confidence interval (CI) for the difference between two population proportions is where
and
are the sample proportion and sample size of the first sample, and
and
are the sample proportion and sample size of the second sample.
We see that and
. We also know that a 98% confidence level corresponds to a critical value of
, so we can plug these values into the formula to get our desired confidence interval:
Hence, we are 98% confident that the true difference in the proportion of people that live in a city who identify as a democrat and the proportion of people that live in a rural area who identify as a democrat is contained within the interval {-0.2941,-0.0337}
The 98% confidence interval also suggests that it may be more likely that identified democrats in a rural area have a greater proportion than identified democrats in a city since the differences in the interval are less than 0.