X is a normally distributed random variable with mean 71 and standard deviation 9.
1 answer:
Since the standard deviation is 10 and the first given score is 40, the 40 is 2 standard deviations below the mean.
On the other hand, 80 is 2 standard deviations above the mean.
Thus, we subtract the probability below the z-score which is 2, 0.97725, and the probability below the z-score which is -2, 0.02275.
We may obtain the values using the z-score table.
Therefore, the approximate value of the probability is 0.955.
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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