Question 1195496: ok, this is a big one. i need help with all of them, please and thank you. i wonder if i could figure this out on wolfram alpha or desmos!
Scores for a common standardized college aptitude test are normally distributed with a mean of 502 and a standard deviation of 111 . Randomly-selected women are given a Preparation Course before taking this test.
a) If one of the women is randomly selected, find the probability that her score is at least 565.7. Enter your answer as a number accurate to 4 decimal places.
b) If nineteen of the women are randomly selected, find the probability that their mean score is at least 565.7. Enter your answer as a number accurate to 4 decimal places.
c) I initially assumed that the Preparation Course had no effect on people's test scores. A random sample of nineteen women does result in a mean score of more than 565.7. Is there strong evidence to support a claim that I was wrong -- and the Preparation Course is actually effective? Yes or No
Yes. The probability indicates that -- if the Preparation Course has no effect -- it is highly unlikely that a randomly selected group of women would get a mean higher than 565.7 by chance.
No. The probability indicates that -- even if the Preparation Course has no effect -- it isn't unlikely to randomly select a group of women with a mean higher than
565.7 by chance.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! for a single sample, your z-score is equal to (565.7 - 502) / 111 = .5738738739, rounded to 10 decimal places.
the probability she will get a z-score less than that is equal to .7169734661.
the probability she will not get a z-score less than that is equal to 1 minus .7169734661 = .2830265339.
that's the probability she will get a z-score greater than or equal to that.
all numbers are rounded to the number of decimal places that can be displayed from the calculator used.
round to 4 decimal places to get .2830.
for a single sample, you use the standard deviation.
for the mean of a sample of size greater than 1, you use the standard error.
if the sample size is 19, then the standard error is equal to the standard deviation divided by the square root of the sample size = 111 / sqrt(19) = 25.46514646.
the z-score is equal to (565.7 - 502) / 25.46514646 = 2.501458223.
the probability she will get a z-score less than that is equal to .9938158335.
the probability that she will not get a z-score less than that is equal to 1 - .9938158335 = .006184665.
that's the probability she will get a z-score greater than or equal to that.
round to 4 decimal places to get .0062.
a probability of .0062 is a very small probability.
that means that it is highly unlikely that a sample of 19 women will have a mean score of 565.7 or greater by chance alone.
that means that, if the mean of the sample of 19 women was 565.7, then something else might be influencing that other than chance.
a likely culprit would be that the preparation course did have an effect on their scores.
i used the ti-84 plus calculator to get these results.
you can get comparable results using the calculator at https://www.calculator.net/z-score-calculator.html
results from using this calculator are:
let me know if you have any questions.
theo
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