The length X of a certain species of tropical fish has a normal distribution wit

Question

The length X of a certain species of tropical fish has a normal distribution with mean equal to 54 mm and standard deviation equal to 4.5 mm. A random sample of four fish was collected from a very large population of such fish and their lengths were measured. a. What is the probability that all four fish selected will have lengths between 51 and 60 mm? b. What is the probability that the mean length of the four fish is between 51 and 60 mm? c. Your answer in part b. should be larger than your answer in part a. Explain why this must be so.

Explanation / Answer

a. From information given, mu=54, and sigma=4.5. Compute z scores for for x1=51 and x2=60 by substituting values in z score formula, z=(x-mu)/sigma, where, x is raw score, mu is population mean, and sigma is population standard deviation.

z1=(51-54)/4.5=-0.67 and z2=(60-54)/4.5=1.33

The two z scores are of opposite signs, therefore, find areas between mean and respective z scores and add them.

P(51<X<60)=0.2486+0.4082=0.6568 (ans)

b. There were four fishes captured, therefore, sample mean, xbar=54, and sample standard deviation, sigma xbar=4.5/sqr 4=2.25.

Obtain the z scores.

z1=(51-54)/2.25=-1.33 and z2=2.67

Therefore, P(51<X<60)=0.4082+0.4962=0.9044 (ans)

c. In part b, it is asked to find probability of mean length of four fishes, that is why sample mean and sample standard deviation was taken into account.

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