SOLUTION
We are given x is normally distributed with mean (µ) = 56 and standard deviation (∂) = 7.
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= 0.2843
Therefore the probability that one catches a fish over 60inches is 0.2843
P(x<50) to calculate this probability we shall calculate the z-value for
x= (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 = -0.86
For p(x<50) we calculate from z tables the area to the left of -0.86 from z tables, thus p(x<50) = 0.1949
Therefore the probability that one catches a fish over 60inches is 0.1949
SOLUTION
Confidence interval ( CI) = Sample Mean (x) ±( (zα * ∂)/√n) where x is sample mean,zα is the value of z for significance level given, ∂ is the standard deviation and √n is the square root of n
Thus, CI = x ± ((zα * ∂)/√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)
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