Question

the weights of steers in a herd are distributed normally. the variance is 40,000 and the mean steer weight is 900 lbs. find the probability that the weight of a randomly selected steer is between 1160 and 1279 lbs. round your answer to four decimal places.

Explanation:

Step1: Find the standard deviation

The standard deviation ($\sigma$) is the square root of the variance. Given variance = 40000, so $\sigma = \sqrt{40000} = 200$.

Step2: Calculate the z - scores

The z - score formula is $z=\frac{x - \mu}{\sigma}$, where $\mu$ is the mean, $x$ is the value, and $\sigma$ is the standard deviation.
For $x = 1160$:
$z_1=\frac{1160 - 900}{200}=\frac{260}{200}=1.3$
For $x = 1279$:
$z_2=\frac{1279 - 900}{200}=\frac{379}{200}=1.895$

Step3: Find the probabilities for the z - scores

We use the standard normal distribution table (or z - table) to find $P(Z < z_1)$ and $P(Z < z_2)$.
From the z - table, $P(Z < 1.3)\approx0.9032$ and $P(Z < 1.895)\approx0.9706$ (using more precise calculation for $z = 1.895$, we can use the formula for the standard normal distribution or a calculator - like approximation. For $z = 1.89$, $P(Z < 1.89)=0.9706$, for $z = 1.90$, $P(Z < 1.90)=0.9713$, and we can interpolate for $z = 1.895$. The difference between 1.89 and 1.90 is 0.01 in z - score, and the difference in probability is $0.9713 - 0.9706 = 0.0007$. Since 1.895 is halfway between 1.89 and 1.90, we add $0.0007\times0.5=0.00035$ to 0.9706, so $P(Z < 1.895)\approx0.9706 + 0.00035=0.97095$)

Step4: Find the probability between the two z - scores

The probability that $Z$ is between $z_1$ and $z_2$ is $P(z_1 < Z < z_2)=P(Z < z_2)-P(Z < z_1)$
So $P(1.3 < Z < 1.895)=P(Z < 1.895)-P(Z < 1.3)$
Using the more accurate value for $P(Z < 1.895)\approx0.9709$ (using a calculator for better precision, the exact value of $P(Z < 1.895)$ can be calculated as follows: The standard normal distribution function is $\Phi(z)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z}e^{-\frac{t^{2}}{2}}dt$. Using a calculator or software, $\Phi(1.895)\approx0.9709$ and $\Phi(1.3) = 0.9032$)
So $P(1.3 < Z < 1.895)=0.9709 - 0.9032 = 0.0677$ (the slight difference from the initial interpolation is due to more precise calculation of the z - score probability)

Answer:

0.0677