Question

The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 1000 lbs. Find the probability that the weight of a randomly selected steer is between 639 and 1420lbs. Round your answer to four decimal places.

Answer 1

Answer:

0.9466 = 94.66% probability that the weight of a randomly selected steer is between 639 and 1420lbs.

Step-by-step explanation:

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

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 1000, \sigma = 200$

Find the probability that the weight of a randomly selected steer is between 639 and 1420lbs.

This is the pvalue of Z when X = 1420 subtracted by the pvalue of Z when X = 639. So

X = 1420

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

$Z = \frac{1420 - 1000}{200}$

$Z = 2.1$

$Z = 2.1$ has a pvalue of 0.9821

X = 639

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

$Z = \frac{639 - 1000}{200}$

$Z = -1.805$

$Z = -1.805$ has a pvalue of 0.0355

0.9821 - 0.0355 = 0.9466

0.9466 = 94.66% probability that the weight of a randomly selected steer is between 639 and 1420lbs.

Answer 2

Answer:

Probability that the weight of a randomly selected steer is between 639 and 1420 lbs is 0.9470.

Step-by-step explanation:

We are given that the weights of steers in a herd are distributed normally. The standard deviation is 200 lbs and the mean steer weight is 1000 lbs.

Let X = weights of steers in a herd

So, X ~ N($\mu=1000,\sigma^{2} = 200^{2}$)

The z score probability distribution is given by;

                 Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean steer = 1000 lbs

            $\sigma$ = standard deviation = 200 lbs

So, probability that the weight of a randomly selected steer is between 639 and 1420 lbs is given by = P(639 lbs < X < 1420 lbs)

  P(639 lbs < X < 1420 lbs) = P(X < 1420 lbs) - P(X $\leq$ 639 lbs)

  P(X < 1420 lbs) = P( $\frac{X-\mu}{\sigma}$ < $\frac{1420-1000}{200}$ ) = P(Z < 2.10) = 0.98214                                                  

  P(X $\leq$ 639 lbs) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{639-1000}{200}$ ) = P(Z $\leq$ -1.81) = 1 - P(Z < 1.81)

                                                            = 1 - 0.96485 = 0.03515

Therefore, P(639 lbs < X < 1420 lbs) = 0.98214 - 0.03515 = 0.9470

Hence, probability that the weight of a randomly selected steer is between 639 and 1420 lbs is 0.9470.