Question

The weights of the fish in a certain lake are normally distributed with a

Answer 1

$P(17.6< xbar < 23.6) = P(-0.8 < t < 1.2 when df = 3) = 0.6730$

Solution

Normal distribution between 17.6 and 23.6

μ  = 20 lb

σ/x  = σ /√n = 9/√9 = 3

Distribution score for  17.6 = 17.6 - 20÷ 3 = - 0.8

$Distribution score for 17.6 = 17.6 - 20÷ 3 = - 0.8$

Distribution score for 23.6 = 23.6 - 20÷ 3 = 1.2

$Distribution score for 23.6 = 23.6 - 20÷ 3 = 1.2$

p(mean weight between 17.6 and 23.6 for 9 fish) =

9(-0.8 ∠≠∠1.2) = 0.8849 - 0,2119 = 0.6730

P(17.6< xbar < 23.6) = P(-0.8 < t < 1.2 when df = 3) = 0.6730

$P(17.6< xbar < 23.6) = P(-0.8 < t < 1.2 when df = 3) = 0.6730$