2 a) p ( x >2 ) ( p z> 25.3 2.1 ) p ( z >1.57 )=10.0582=0.9418 b) P ( Z > z )=0.90 P ( Z > z )=0.10 z=1.28= x5.3 2.1 x=7.988 Problem 3 a) =64 =3 For 1st quartile, P ( X x )=0.25 or z=0.675 ( P z x64 =0.25 3 ) P ( Z z )=0.25 in Z score =0.675 ( x64 3 ) x=(0.675 x 3 ) +64=61.975 62 Therefore 1st quartile = 62 inches b) For 90th percentile, P ( X x )=0.9 or P ( Z z )=0.9 in Z score z=1.2816 ( P z x64 =0.9 3 ) =1.2816 ( x64 3 ) x=( 1.2816 x 3 ) +64=67.8448 68 Therefore 90th percentile = 68 inches c) P ( x >70 ) =1P( x 70) ( P ( x >70 ) =1P z 7064 =1 p(z 2) 3 ) P ( x >70 ) =1P ( z 2 )=10.97725=0.02275 d) n=100 Std Error of mean= 3 = =0.3 n 100 e) P ( X <68 )= 6864 > 0.5 b) Hypothesized po 0.5 std error of proportion= po (1 po) =0.040824829 n c) x Point estimate of p , pv = =0.553333333 n Z statistic z= p v po =1.306394529 standard errot of proportion d) This is a 1 tailed test hence p=0.095709213 e) Since p < 0.1, we FAIL TO REJECT THE NULL HYPOTHESIS No significance evidence that there are more male than female students Problem 7 Group 1 (Before) 1200 1150 Group 2 (After) 1260 1120 Difference 60 -30 1260 1050 980 1260 1200 1050 0 150 70 a) Formulating hypothesis H 0 :=2 1 0 H a :=2 1> 0 b) We want to evaluate the difference between the test scores before and after the course c) Significance level=0.1 From observation this is a right tailed test St andard deviation of thedifference , s=49.25967064 Standard error of the difference , sD= =22.02959442 n Mean of difference , XD=50 At t = [XD -uD]/sD, where uD = the hypothesized difference = 0 then t=2.269674105 d) As df =n1=4 P=0.042879254 e) Also comparing this p to the significance level, we REJECT the NULL HYPOTHESIS Therefore there is sufficient evidence to support the claim that taking a preparatory course increases an SAT test score. Problem 8 a) Formulating hypothesis H 0 :=2 1 0 H a :=2 1> 0 b) Significance level=0.1 From observation this is a left tailed test Calculating means X 1=18 X 2=6 Calculating standard deviations s 1=20 s 2=8 s s ( 1)2 ( 2)2 + n1 n2 standard error of their difference , sD= n1=sample group1=100 n2=sample group2=130 Thus df =n1 +n22=228 sD=2.119506474 Therefore the t statistic will be t= [ X 1 X 2uD] =5.661695375 sD c) uD = 0 Using p values as df = 228 P=2.24233 x 10 8 Comparing to significance level, we REJECT the NULL HYPOTHESIS This shows there is sufficient evidence that that mean age of entering the workforce in Canada is lower than the mean age in the U.S