Alivestock 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?

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 and standard deviation , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

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Item a:

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

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:

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

has a p-value of 0.9463.

X = 1150

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.

A similar problem is given at link

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.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14