Question

the chickens of the ornithes farm are processed when they are 20 weeks old. the distribution of their weights is normal with a mean of 3.8 lb. and a standard deviation of 0.6 lb. On your paper draw a normal curve for each situation and write what you typed in the calculator. Answer the following.

a. What proportion of chickens weigh less than 3 pounds? (Round to 4 decimal places.)

b. What proportion of chickens weigh between 3.5 and 4.5 pounds? (Round to 4 decimal places.)

c. What proportion of chickens weigh more than 5 pounds? (Round to 4 decimal places.)

d. 20 percent of the chickens have a weight below what value? (Round to 1 decimal place.)

e. If I randomly select 5 chickens, what is the probability that they have weights with a mean more than 4 pounds? (Round to 4 decimal places)

Answer 1

Solution :

Given that,

mean = = 3.8

standard deviation = = 0.6

a ) P( x < 3 )

P ( x - / ) < ( 3 - 3.8 / 0.6)

P ( z < -0.8 / 0.6 )

P ( z < -0.13)

= 0.4483

Probability =0.4483

b ) P (3.5 < x < 4.5 )

P ( 3.5 - 3.8 / 0.6) < ( x -  / ) < ( 4.5 - 3.8 / 0.6)

P ( - 0.3 / 0.6 < z < 0.7 / 0.6 )

P (-0.5< z < 1.17)

P ( z < 1.17 ) - P ( z < -0.5)

Using z table

= 0.8790 - 0.3085

= 0.5705

Probability = 0.5705

c ) P (x > 5 )

= 1 - P (x < 5)

= 1 - P ( x -  / ) < ( 5 - 3.8 / 0.6)

= 1 - P ( z < 1.2 / 0.6 )

= 1 - P ( z < 2)

Using z table

= 1 - 0.9772

= 0.0228

Probability = 0.0228

d ) Using standard normal table,

P(Z < z) = 20%

P(Z < z) = 0.20

P(Z < - 0.61) = 0.20

z = - 0.61

Using z-score formula,

x = z * +

x = -0.84 * 0.6 + 3.8

x = 3.3

e ) n = 5

= 3.8

= / n = 0.6 5= 37.94733

P ( > 5 )

= 1 - P ( < 5)

= 1 - P ( - / ) < ( 4 - 3.8 / 0.2683)

= 1 - P ( z < 0.2 / 0.2683 )

= 1 - P ( z < 0.74)

Using z table

= 1 - 0.7704

= 0.2296

Probability = 0.2296