Question

Weights of Northern Pike follow a normal distribution with mean weight of 1.32 pounds and a standard deviation of .6 pounds.

(A). Find the probability a randomly selected pike weights less than 2 pounds.

(B). Find the probability a randomly selected pike weights more than 1 pound?

(C) Find the probability a randomly selected pike weights between 1.2 and 2.2 pounds.

(D)What pike weight corresponds to the 67th percentile?

(E) Suppose you find a random sample of n=16 pike, what is the probability that the sample mean will fall between 1.2 and 2.2 pounds?

Answer 1

Solution :

Given that ,

mean = = 1.32

standard deviation = =.6

(A).

P(x < 2) = P[(x - ) / < (2 - 1.32) / .6]

= P(z < 1.13)

= 0.8708

Probability = 0.8708

(B).

P(x > 1) = 1 - P(x < 1)

= 1 - P[(x - ) / < (1 - 1.32) / .6)

= 1 - P(z < -0.53)

= 1 - 0.2981

= 0.7019

Probability = 0.7019

(C).

P(1.2 < x < 2.2) = P[(1.2 - 1.32)/ .6) < (x - ) /  < (2.2 - 1.32) / .6) ]

= P(-0.2 < z < 1.47)

= P(z < 1.47) - P(z < -0.2)

= 0.9292 - 0.4207

= 0.5085

Probability = 0.5085

(D).

Using standard normal table ,

P(Z < z) = 67%

P(Z < 0.44) = 0.67

z = 0.44

Using z-score formula,

x = z * +

x = 0.44 * .6 + 1.32 = 1.58

The pike weight corresponds to the 67th percentile is 1.58

(E).

= / n = .6/ 16 = 0.15

= P[(1.2 - 1.32) /0.15 < ( - ) / < (2.2 - 1.32) / 0.15)]

= P(-0.8 < Z < 5.87)

= P(Z < 5.87) - P(Z < -0.8)

= 1 - 0.2119

= 0.7881

Probability = 0.7881