5. The owner of a fish market determined that the average weight for a catfish i

Q: 5. The owner of a fish market determined that the average weight for a catfish i

a. Since =3.6 and =0.8 we have: P ( X>4.8 )=P ( X>4.83.6 )=P ( X>4.83.60.8) Since Z=x and 4.83.60.8=1.5 we have: P ( X>4.8 )=P ( Z>1.5 ) Use the standard normal table to conclude that: P (Z>1.5)=0.0668 b. Since =3.6 and =0.8 we have: P ( 3

ID: 3274350 • Letter: 5

Question

5. The owner of a fish market determined that the average weight for a catfish is 3.6 pounds with a standard deviation of 0.8 pound. Assume the weights of catfish are normally distributed.

PLEASE SHOW WORK AND FORMULAS USED. EXCEL, ETC IF NECESSARY

a. What is the probability that a randomly selected catfish will weigh more than 4.8 pounds?

b. What is the probability that a randomly selected catfish will weigh between 3 and 5 pounds?

c. A randomly selected catfish will weigh more than x pounds to be one of the top 5% in weight. What is the value of x?

d. A randomly selected catfish will weigh less than x pounds to be one of the bottom 20% in weight. What is the value of x?

e. Above what weight in (pounds) do 87.70% of the weights occur?

f. What is the probability that a randomly selected catfish will weigh less than 3.2 pounds?

g. Below what weight (in pounds) do 83.4% of the weights occur?

Explanation / Answer

Since =3.6 and =0.8 we have:

P ( X>4.8 )=P ( X>4.83.6 )=P ( X>4.83.60.8)

Since Z=x and 4.83.60.8=1.5 we have:

P ( X>4.8 )=P ( Z>1.5 )

Use the standard normal table to conclude that:

P (Z>1.5)=0.0668

Since =3.6 and =0.8 we have:

P ( 3<X<5 )=P ( 33.6< X<53.6 )=P ( 33.60.8<X<53.60.8)

Since Z=x , 33.60.8=0.75 and 53.60.8=1.75 we have:

P ( 3<X<5 )=P ( 0.75<Z<1.75 )

Use the standard normal table to conclude that:

P ( 0.75<Z<1.75 )=0.7333

c. Top 5%

z from table =1.645

(x-3.6)/0.8 = 1.645

Solve for x

= 4.916

z from table =0.84

(x-3.6)/0.8 = 0.84

Solve for x

= 2.927

87.7% of the weights

z from table =1.21

(x-3.6)/0.8 = 1.21

Solve for x

= 4.569

Since =3.6 and =0.8 we have:

P ( X<3.2 )=P ( X<3.23.6 )=P (X<3.23.60.8)

Since x=Z and 3.23.60.8=0.5 we have:

P (X<3.2)=P (Z<0.5)

Use the standard normal table to conclude that:

P (Z<0.5)=0.3085

g. below 83.4%

z from table =0.97

(x-3.6)/0.8 = 0.97

Solve for x

= 4.376

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5. The owner of a fish market determined that the average weight for a catfish i

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