Question

A livestock company reports that the mean weight of a group of young steers is 1143 pounds with a standard deviation of 88 pounds. Based on the model ​N(1143​,88​) for the weights of​ steers, what percent of steers weigh

-Under 1200
-Between 1100 and 1250 pounds?

Answer 1

Answer:

a) P(x<1200)=74.14%

b) P(1100<X<1250)=57.54%

Step-by-step explanation:

Normal Distribution

The normal distribution, also known as the bell curve, is a distribution that occurs naturally in many situations of life. We use the model $N(\mu,\sigma)$ to understand the behavior of some real-life variables. Where $\mu$ is the mean value and $\sigma$ is the standard deviation.

In our case, we have

$\mu=1143,\ \sigma=88$

And are required to find the percentage of steers whose weigh lie within a given range. We must use some sort of table or digital means to compute the values because the normal distribution cannot be calculated directly by a formula. We use the NORMDIST (or NORM.DIST) formula for Excel which gives us the left tail of the area behind the bell curve, i.e. the cumulative percentage for a give value of X. The formula has the form

NORM.DIST(x,mean,standard_dev,cumulative)

a) X<1200

The formula is used with the following parameters

NORM.DIST(1200,1143,88,true)

and we get

$P(X<1200)=0.7414=74.14\%$

b) We need to compute P(1100<X<1250). To do this, we calculate both left tails and the subtract them

NORM.DIST(1100,1143,88,true)=0.3125

NORM.DIST(1250,1143,88,true)=0.8880

$P(1100<X<1250)=0.8880-0.3125=0.5754$

$\boxed{P(1100<X<1250)=57.54\%}$