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Questions (9)
If weight in the general population is normally distributed with an average of 160 and a standard deviation of 20 pounds, what
4 answers
6,159 views
Calculate the Pearson product-moment correlation for the data below.
X 3 4 2 1 Y 5 5 3 4
1 answer
1,931 views
If weight in the general population is normally distributed with an average of 160 and a standard deviation of 20 pounds, what
0 answers
520 views
Students in the psychology department consume an average of 5 cups of coffee per day with a standard deviation of 1.75 cups. The
2 answers
4,387 views
For a hypothetical normal distribution of test scores, approximately 95% fall between 38 and 62, 2.5% are below 38, and 2.5% are
2 answers
1,239 views
The producers of a new toothpaste claim that it prevents more cavities than other brands of toothpaste. A random sample of 60
1 answer
2,554 views
What is the probability of pulling a heart from a standard (52-card) deck of playing cards?
A. .25 B. .50 C. .75 D. 1.0 Thanks
2 answers
1,209 views
Faculty in the psychology department at APUS consume an average of 5 cups of coffee per day with a standard deviation of 1.5.
1 answer
932 views
In a psychology class of 100 students, test scores are normally distributed with a mean of 80 and a standard deviation of 5.
3 answers
2,445 views
Answers (3)
z = (6 - 5)/1.75 = 0.57 .2157+.50= 0.7157x100=71.57% i.e 71.57% In the second part, find z using a z-table for the 80th percentile. Plug into the formula, along with mean and standard deviation. Solve for x. z = (x - mean)/sd z (sd)= X-mean z (sd)+mean= X
Thanks. For the z scores I have: z = (120 - 160)/20 = -40/20=-2 z = (170 - 160)/20 = 10/20= 0.5 and the z table for 2 -(area between Mean and z) 0.47725, area beyond: 0.02275 For 0.5: 0.19146 and area beyond: 0.30854 What do I do after that, and which # I
Thank you very much! I am having real problems with that. Is that the formula that should be used: ơ=√∑ (x- µ)2/N ?
