1. Youre going to catch a fish in the river. You know that its size is normally distributed with mean 56 inches and standard deviation 7 inches. What is the probability that you catch a fish over 60 inches? What is the probability that you catch a fish less than 50 inches?SOLUTION We are given x is normally distributed with mean (µ) = 56 and standard deviation (?) = 7.a) P(x> 60) To calculate this probability we shall calculate the z-value for x= (60) then we calculate the area to the right of this value from the z-test tables.Z-value is given by, z = (x – µ)/(?) where x is sample statistic,µ is the population mean and ? is the standard deviation.Z for x = 60, z(60) = (60-56)/7 = 0.5714 to2decimal place is 0.57 For p(x>60) we calculate from z tables the area to the right of 0.57 from z tables, thus p(x>60) = 1-0.7157= 0.2843Therefore the probability that one catches a fish over 60inches is 0.2843 b) We are given x is normally distributed with mean (µ) = 56 and standard deviation (?) = 7.P(x<50) to calculate this probability we shall calculate the z-value forx= (57) then we calculate the area to the left of this value from the z-test tables.Z-value is given by, z = (x – µ) / (?) where x is sample statistic,µ is the population mean and ? is the standard deviation.Z for x = 50, z (50) = (50-56)/7 = -6/7 = -0.8571 to 2decimal place = -0.86For p(x<50) we calculate from z tables the area to the left of -0.86 from z tables, thus p(x<50) = 0.1949Therefore the probability that one catches a fish over 60inches is 0.1949 2. Construct a 95% confidence interval if the sample size is 30, mean is 60 and the standard deviation is 8.SOLUTIONConfidence interval ( CI) = Sample Mean (x) ±( (z? * ?)/?n) where x is sample mean,z? is the value of z for significance level given, ? is the standard deviation and ?n is the square root of nThus, CI = x ± ((z? * ?)/?n),CI = 60 ± ((Z0.05 * 8) / ?30), = 60 ± ((1.96 * 8) / ?30 CI = (60 – 2.8628, 60 – 2.8628)CI = (57.1372, 62.8628)Therefore the 95%confidence interval is (57.1372, 62.8628)Hide