deviation

 1. You’re 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 o

1.      You’re 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= 0.2843

Therefore 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 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

2.      Construct a 95% confidence interval if the sample size is 30, mean is 60 and the standard deviation is 8.

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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