Question

the mean weight of a breed of yearling cattle is 1093 pounds. suppose that weights of all such animals can be described by a normal model with a standard deviation of 88 pounds. a) how many standard deviations from the mean would a steer weighing 1000 pounds be? b) which would be more unusual, a steer weighing 1000 pounds, or one weighing 1250 pounds? a) a steer weighing 1000 pounds is standard deviations below the mean. (round to two decimal places as needed.)

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. We want to find the number of standard deviations $x = 1000$ is from the mean $\mu=1093$ with $\sigma = 88$.

Step2: Calculate the z - score

Substitute the values into the formula: $z=\frac{1000 - 1093}{88}=\frac{- 93}{88}\approx - 1.06$.

Step3: Interpret the result

The negative sign indicates that 1000 is below the mean. The absolute value of the z - score gives the number of standard deviations from the mean. So 1000 pounds is approximately 1.06 standard deviations below the mean.

Step4: Determine unusualness

For part b, in a normal distribution, values with $|z|>2$ are often considered unusual. For $x = 1000$, $|z|\approx1.06<2$, and for $x = 1250$, $z=\frac{1250 - 1093}{88}=\frac{157}{88}\approx1.78<2$. So neither is very unusual.

Answer:

a. - 1.06
b. Neither a steer weighing 1000 pounds nor one weighing 1250 pounds is very unusual.