# Essay about Unit 4 Problem Set 1: Normal Probability Distributions

673 Words Apr 22nd, 2014 3 Pages
4/27/2014
MA3110
Statistics
Otis Jackson
Unit 4 problem set 1: Normal Probability Distributions
Page.285 Ex 6,8,10,12
6. x = 80, z=80-10015 = -1.33 z= 0.0918 1-0.0918 = 0.9082
8. x = 110, z=110-10015 = 0.67 z= 0.7486 z= 75-10015 = -1.67 z= 0.0475 0.7486-0.0475= 0.7011 (shaded area)
10. z= 0.84 (shaded) z= -0.84 x= 100+(-0.84∙15) = 87 (rounded)
12. . z= 2.33 x= 100+(2.33∙15) = 135 (rounded)
Page 288 Ex 34
34.Appendix B Data Set: Duration of Shuttle Flights
a. Find the mean and standard deviation, and verify that the data have a distribution that is roughly normal. Mean= 25317115 = 220.15 Standard Deviation=115253172-(25317)2115(115-1) = 86 (rounded)
The normal distribution is 115
b. Treat the
…show more content…
c) Population is 25 = .4 Yes the sample distribution of proportion is used when estimating the population proportion.
Page310 ex 14
Being that the questions are starting to become like a small paragraph im just going to but my answers from here on out…
A. mean = 6.0 in. Standard Deviation= 1.0 in. (6.2-6.0)/1 = .20 z score = .5793
B. n=100 1.0100 = 0.1 z = 6.2-6.00.1 = 2 z score=.9772 c) 1 - .5793 = .4207 So 42.07% of men has a head breadth of more than 6.2 inches.
Page 321 Ex 28
. n = 784 p = 7.9% q = 1 - .079 = .921 np = 784(.079) = 61.936 ≥ 5 nq = 784(.921) = 722.064 ≥ 5 σ = 784.079(.921) = 7.55 µ = np = 784(.079) = 61.936 P(x > 478.5) z = 478.5-61.9367.55 = 55.17 1 - .9999 = .0001
Not even going to lie got help with this but apparently Yes and that’s because 99.99% has a normal pattern and being that the other checks fall in the .ooo1% its highly probable that those checks from the other companies are faker that a 4 dollar bill.
Page 329 6,8
6. Yes, the plots follow the line. Which means the plot shows a population with a normal distribution
8. No, the plots do not follow the line
Page330 10
10.

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