Question

A livestock company reports that the mean weight of a group of young steers is 1100 pounds with a standard deviation of 93 pounds. Based on the model ​N(1100​,93​) for the weights of​ steers, what percent of steers weigh ​a) over 1050 ​pounds? ​b) under 1300 ​pounds? ​c) between 1150 and 1250 ​pounds?

Answer 1

Explanation:

a) Calculate the z-score.

z = (x − μ) / σ

z = (1050 − 1100) / 93

z = -0.54

From a z-score table:

P(z<-0.54) = 0.2946

Therefore:

P(z>-0.54) = 1 - 0.2946

P(z>-0.54) = 0.7054

70.54% of steers are over 1050 pounds.

b) Calculate the z-score.

z = (x − μ) / σ

z = (1300 − 1100) / 93

z = 2.15

From a z-score table:

P(z<2.15) = 0.9842

98.42% of steers are under 1300 pounds.

c) Calculate the z-scores.

z = (x − μ) / σ

z = (1150 − 1100) / 93

z = 0.54

z = (1250 − 1100) / 93

z = 1.61

From a z-score table:

P(z<0.54) = 0.7054

P(z<1.61) = 0.9463

Therefore:

P(0.54<z<1.61) = P(z<1.61) - P(z<0.54)

P(0.54<z<1.61) = 0.9463 - 0.7054

P(0.54<z<1.61) = 0.2409

24.09% of steers weigh between 1150 pounds and 1250 pounds.

Answer 2

Using the normal distribution, we find that:

a) 70.54% of steers weigh over 1050 pounds.

b) 1.58% of steers weigh under 1300 pounds.

c) 24.09% of steers weigh between 1150 and 1250 pounds.

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

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• Mean of 1100 means that $\mu = 1100$
• Standard deviation of 93 means that $\sigma = 93$

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Item a:

The proportion over 1050 pounds is 1 subtracted by the p-value of Z when X = 1050, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1050 - 1100}{93}$

$Z = -0.54$

$Z = -0.54$ has a p-value of 0.2946.

1 - 0.2946 = 0.7054.

0.7054 x 100% = 70.54% of steers weigh over 1050 pounds.

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Item b:

The proportion under 1300 pounds is the p-value of Z when X = 1300, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1300 - 1100}{93}$

$Z = 2.15$

$Z = 2.15$ has a p-value of 0.9842.

1 - 0.9842 = 0.0158

0.0158 x 100% = 1.58%

1.58% of steers weigh under 1300 pounds.

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Item c:

The proportion between 1150 and 1250 pounds is the p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1150, thus:

X = 1250

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1250 - 1100}{93}$

$Z = 1.61$

$Z = 1.61$ has a p-value of 0.9463.

X = 1150

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1150 - 1100}{93}$

$Z = 0.54$

$Z = 0.54$ has a p-value of 0.7054.

0.9463 - 0.7054 = 0.2409

0.2409 x 100% = 24.09%

24.09% of steers weigh between 1150 and 1250 pounds.

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