The mean weight of a breed of yearling cattle is 11921192 pounds. Suppose that weights of all such animals can be described by the Normal model N(11921192,7878). a) How many standard deviations from the mean would a steer weighing 10001000 pounds be? b) Which would be more unusual, a steer weighing 10001000 pounds, or one weighing 12501250 pounds?

a) A steer weighing 10001000 pounds is 2.462.46 standard deviations below the mean. (Round to two decimal places as needed.)
b) The steer weighing 1000 pounds is more unusual.

a) 2.46 standard deviations below the mean

b) The steer weighing 1000 pounds is more unusual.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

a) How many standard deviations from the mean would a steer weighing 1000 pounds be?

This is Z when X = 1000.

is 2.46 standard deviations below the mean.

b) Which would be more unusual, a steer weighing 1000 pounds, or one weighing 1250 pounds?

The higher the absolute value of the z-score, the more unusual the measure is.

X = 1000 pounds

The absolute value of -2.46 is 2.46

X = 1250 pounds

The absolute value of 0.74 is 0.74.

So the steer weighing 1000 pounds is more unusual.

2 5/8 is the final answer! : )

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step-by-step explanation: