Question

The weights of steers in a herd are distributed normally. The variance is 10,000 and the mean steer weight is 1400lbs. Find the probability that the weight of a randomly selected steer is between 1539 and 1580lbs. Round your answer to four decimal places.

Answer 1

Answer:

$P(1539<X<1580)=P(\frac{1539-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{1580-\mu}{\sigma})=P(\frac{1539-1400}{100}<Z<\frac{1580-1400}{100})=P(1.39<z<1.8)$

And we can find this probability using the normal standard table with this difference:

$P(1.39<z<1.80)=P(z<1.80)-P(z<1.39)= 0.9641-0.9177=0.0464$

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(1539,1580)$  

Where $\mu=1400$ and $\sigma=\sqrt{10000}= 100$

We are interested on this probability

$P(1539<X<1580)$

And we can solve the problem using the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

Using this formula we got:

$P(1539<X<1580)=P(\frac{1539-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{1580-\mu}{\sigma})=P(\frac{1539-1400}{100}<Z<\frac{1580-1400}{100})=P(1.39<z<1.8)$

And we can find this probability using the normal standard table with this difference:

$P(1.39<z<1.80)=P(z<1.80)-P(z<1.39)= 0.9641-0.9177=0.0464$