Question #272489

The population of fish in a particular pond is known to have a mean length µ = 15 cm with standard deviation σ = 5. You catch fifty fish from the pond and measure their lengths. What is the probability that the average length of those fish (i.e. sample mean X ) is between 14 cm and 16.5 cm?


1
Expert's answer
2021-11-30T10:32:47-0500

Let X=X= sample mean: XN(μ,σ2/n).X\sim N(\mu, \sigma^2/n).

Given μ=15,σ=5,n=50.\mu=15, \sigma=5, n=50.


P(14<X<16.5)=P(X<16.5)P(X14)P(14<X<16.5)=P(X<16.5)-P(X\leq 14)

=P(Z<16.5155/50)P(X14155/50)=P(Z<\dfrac{16.5-15}{5/\sqrt{50}})-P(X\leq \dfrac{14-15}{5/\sqrt{50}})

P(Z<2.1213)P(X1.4142)\approx P(Z<2.1213)-P(X\leq -1.4142)

0.983050.078650.9044\approx0.98305-0.07865\approx 0.9044

The probability that the average length of those fish (i.e. sample mean X ) is between 14 cm and 16.5 cm is 0.9044.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS