Question

Chickens at a farm are processed when they are 20 weeks old. The distribution of their weights is normal with mean 1.7 kg, and standard deviation 0.250kg. The farm has created three categories for these chickens according to their weight: Petite (weight less than 1.5 kg), Standard (weight between 1.5kg and 2.2kg), and Big (weight above 2.2kg).

1)

(a)-What proportion of these chickens will be in each category?

(b)-Draw a normal distribution graph and sow these proportions in different colors.

2) What weight level (L in kg) is exceeded by 60% of the population?

3) Let a and b, (a < b) be two weight values symmetric around the mean (1.7 kg), (i.e., the mean is at the center of the interval [a, b]), find the values of a and b between which lies 90% of the population.

4) Assume that the mean weight of chickens can be adjusted easily, but the standard deviation remains the same (0.250 kg), at what value the mean should be set so that only 10% of the chickens are Petite.

5) Suppose that 4 chickens are selected at random. What is the probability that one of them will be petite?

Answer 1

The distribution of chicken is normal with mean ()= 1.7 kg and sd = 0.25 kg

The three categories of chicken are

Petitte: with weight <1.5 kg

Standard: with weight 1.5 kg to 2.2 kg

Big: with weight > 2.2 kg

The Z value for Petite is

P(Z<1.5) = Area under the standard normal curve below -0.8 = Area of 0.8 to infinity = 0.5 - Area between 0 to 0.8

= 0.5 - 0.2881=0.2119

The two Z Scores of standard variety are

respectively

P(1.5<Z<2.2) = Area under the standard normal curve between -0.8 to 2

= Area under 0 to 0.8 + Area under 0 to 2

= 0.2881 + 0.4772 = 0.7653

and the Z score of Big variety is

P(Z>2.2) = Area under the standard normal curve more than 2

= Area under the standard normal curve from 0 to infinity - area of the standard normal curve between 0 to 2

= 0.5 - 0.4772 = 0.0228

So the required proportion of Petitte, Standard and Big are 21.19%, 76.53% and 2.28% respectively

2. Weight Level in Kg which exceeds 60% of the population

We need to find L such that

P(X>L) = 0.60

Therefore we need to find the Z value such that are under the standard normal curve between Z to infinity = 0.6

Or, Area between 0 to infinity - Area between 0 to Z = 0.6

or, 0.5 - Area between 0 to Z = 0.6

or,

or, kg

3. We need to find out a and b such that. It is given a and b are equidistant from 0 in either side

P(a<X<b) = 0.9

Therefore we need to find the

Area between 0 to a + Area between 0 to b

Since a and b are equidistant from 0, we need to find

2 * area between 0 to b such that it is equal to 0.9

So Area between 0 to be is equal to 0.9/2 = 0.45

or,

Therefore the value of a and b are -1.8125 and 1.8125 respectively

4. We need to find out the mean such that 10% of the chicken will be petite

The Z score

The area under the standard normal curve should be 0.10

Therefore we need to fin the Z value such that left side of Z value should be equal to 0.10.

Using similarity of standard normal curve, this is also the right side of Z value between Z to infinity

= Area between 0 to infinity - area between 0 to Z

= 0.5 - area between 0 to Z

So we have the equation

Solving the same, we get the new mean as 1.4

5. Probability that a chicken is petite = p = 0.2119

Probability that the chicken is not petite = q = 1 - 0.2119 =  0.7881

4 chickens were drawn = n = 4

So the distribution follows Binomial with success p

The distribution is given as

Therefore