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Oduvanchick [21]
3 years ago
9

The weights of steers in a herd are distributed normally. The standard deviation is

Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

0.9317

Step-by-step explanation:

Standard deviation of the weights = \sigma=100 lbs

Mean weight = u = 1200 lbs

We need to find the probability that the weight(x) of a randomly selected steer is between 1000 lbs and 1369 lbs i.e. P(1000 < x < 1369)

Since, weights follow the normal distribution we can use the z values to find the required weight. For this we have to convert both the values to z score. The formula for z scores is:

z=\frac{x-u}{\sigma}

1000 converted to z scores is:

z=\frac{1000-1200}{100}=-2

1369 converted to z scores is:

z=\frac{1369-1200}{100}=1.69

So, we have to find the values from z table that lie between -2 to 1.69

P( 1000 < x < 1369 ) = P(-2 < z < 1.69)

P(-2 < z < 1.69) = P(z < 1.69) - P(z < -2)

From the z table:

P(z < 1.69) = 0.9545

P(z < -2) = 0.0228

So,

P(-2 < z < 1.69) = 0.9545 - 0.0228 = 0.9317

Thus,

P( 1000 < x < 1369 ) = 0.9317

From this we can conclude that:

The probability that the weight of a randomly selected steer is between 1000 lbs and 1369 lbs is 0.9317

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