Question

Please help me find the solutions with the work. I’ve done the work and the answers are in the back but my answers don’t correlate and I wanna see where I might have gone wrong.

Statistics Cumulative Final Exam Review 1) For the population: 15 22 17 18 28 find the mean and standard deviation 2) Find the mean and standard deviation for the following frequency distribution Class 0- 4 5-9 Ereauency 10-14 15-19 20-24 3) Based on experience, Alice believes the probabilities that it wil tke her 1, 2, 3, or 4 hours to complete her homework in an evening are .21, .34, 29, and 16 respectively. How long can Alice expect to complete her homework in an evening and what is the standard deviation on the number of hours she will spend on homework? The Chamber of Commerce is checking of small businesses are behind on their loans. They took a random sample of 25 local businesses to determine how many are behind on their loans. Find the probability that 4) credit problems of small businesses. They have found that 20% a) exactly 6 local businesses are behind on their loans (binomial formula) b) less than 6 local businesses are behind on their loans (table) d) How many local businesses can be expected to be behind on their loans? What is the standard deviation? The average number of pages of a statistics book that a student can read in an hour is 7, with a standard deviation of 1.5 and the population is normally distributed. What is the probability that a randomly selected student can read 5) a) more than 11.5 pages in an hour? b) between 5 and 11.5 pages in an hour?

Answer 1

$1.~Mean=(15+22+17+18+28)/5=20\\ \sum x^2=(15^2+22^2+17^2+18^2+28^2)=2106\\ \standard~deviation=\sqrt{(\sum x^2-5\bar{x}^2)/5}=4.6043\\ 2.~$

$Mean=\bar{x}=10.5\\ Standard~deviation=\sqrt{\frac{1}{19}(\sum x^2f -20*\bar{x}^2)}\\ =\sqrt{(2960-20*10.5^2)/19}=6.3037\\ 3.~E(X)=\mu=1*0.21&plus;2*0.34&plus;3*0.29&plus;4*0.16= 2.4\\ E(X^2)=1^2*0.21&plus;2^2*0.34&plus;3^2*0.29&plus;4^2*0.16= 6.74\\ Var(X)=\sigma^2=E(X^2)-E^2(X)=6.74-2.4^2=0.98\\ \sigma=\sqrt{0.98}=0.9899\\ 4.~Let~X~be~a~random~variable~denoting~no.~of~local~business~behind~on~their~loan.\\ Then~X\sim Binomial(25,~0.2).\\ (a)~P(X=6)=\binom{25}{6}(0.2)^6(1-0.2)^{25-6}=0.1633\\ (b)~P(X<6)=\sum_{x=0}^5\binom{25}{x}(0.2)^x(1-0.2)^{25-x}=0.6167\\ (c)~E(X)=25*0.2=5\\ Var(X)=25*0.2*0.8=4\\ standard~deviation=2\\$

$5.~a)~P(X>11.5)=1-\Phi((11.5-7)/1.5)=0.0013\\ b)~P(5<X<11.5)=\Phi((11.5-7)/1.5)-\Phi((5-7)/1.5)=0.9074\\$