Question

1) A random variable is normally distributed with a mean of 50 and a standard deviation of 5.

a. What is the probability the random variable will assume a value between 45 and 55? b. What is the probability the random variable will assume a value between 40 and 60?

2) Golden Harvest produces high-quality organic frozen turkeys for distribution in organic food markets. The company has developed a range feeding program with organic grain supplements to produce their product. The mean weight of its frozen turkeys is 15 pounds with a variance of 4. Historical experience indicates that weights can be approximated by the normal probability distribution. Market research indicates that sales for frozen turkeys over 18 pounds are limited. a. What is the probability of the company’s turkey units will be over 18 pounds? b. What is the percentage of the company’s turkey units will be over 25 pounds?

Answer 1

1.) Here

Mean $\mu =50$

SD $\sigma = 5$

(a) Between 45 and 55

$z_1=\dfrac{\bar x_1-\mu}{\sigma}=\dfrac{45-50}{5}=-1$

$z_2=\dfrac{\bar x_2-\mu}{\sigma}=\dfrac{55-50}{5}=1$

So, $P(-1<z<1)=0.6827$

(b) between 40 and 60

$z_1=\dfrac{\bar x_1-\mu}{\sigma}=\dfrac{40-50}{5}=-2$

$z_2=\dfrac{\bar x_2-\mu}{\sigma}=\dfrac{60-50}{5}=2$

So, $P(-2<z<2)=0.9545$

2.) The distribution given here is,

X ~ N($\mu=15,\sigma =2$ )

The required probability here is :

(a) P(X > 18)

Converting into a standard normal variable, we get

$P(Z>\dfrac{18-15}{2})=P(Z>1.5)$

Getting it from the standard normal tables, we get:

$P(Z>1.5)=0.0668$

(b) The required probability here is

P(X > 25)

Converting into a standard normal variable, we get

$P(Z>\dfrac{25-15}{2})=P(Z>5)$

Getting it from the standard normal tables, we get:

$P(Z>5)\approx 0$

Therefore 0% of the company's turkey units will be over 25 pounds