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Brilliant_brown [7]
3 years ago
10

A livestock company reports that the mean weight of a group of young steers is 1143 pounds with a standard deviation of 88 pound

s. Based on the model ​N(1143​,88​) for the weights of​ steers, what percent of steers weigh
-Under 1200
-Between 1100 and 1250 pounds?
Mathematics
1 answer:
malfutka [58]3 years ago
6 0

Answer:

<em>a) P(x<1200)=74.14%</em>

<em>b) P(1100<X<1250)=57.54%</em>

Step-by-step explanation:

<u>Normal Distribution</u>

The normal distribution, also known as the bell curve, is a distribution that occurs naturally in many situations of life. We use the model N(\mu,\sigma) to understand the behavior of some real-life variables. Where \mu is the mean value and \sigma is the standard deviation.

In our case, we have

\mu=1143,\ \sigma=88

And are required to find the percentage of steers whose weigh lie within a given range. We must use some sort of table or digital means to compute the values because the normal distribution cannot be calculated directly by a formula. We use the NORMDIST (or NORM.DIST) formula for Excel which gives us the left tail of the area behind the bell curve, i.e. the cumulative percentage for a give value of X. The formula has the form

NORM.DIST(x,mean,standard_dev,cumulative)

a) X<1200

The formula is used with the following parameters

NORM.DIST(1200,1143,88,true)

and we get

P(X

b) We need to compute P(1100<X<1250). To do this, we calculate both left tails and the subtract them

NORM.DIST(1100,1143,88,true)=0.3125

NORM.DIST(1250,1143,88,true)=0.8880

P(1100

\boxed{P(1100

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