Question

The weights of steers in a herd are distributed normally. The variance is 40,00040,000 and the mean steer weight is 1400lbs1400 lbs. Find the probability that the weight of a randomly selected steer is between 17391739 and 1800lbs1800 lbs. Round your answer to four decimal places.

Answer 1

Answer:

The probability that the weight of a randomly selected steer is between 1,739 and 1,800 lbs is 0.0218 or 2.18%

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

μ of the weights of steers in a herd= 1,400 lbs

σ² = 40,000 ⇒ σ = √40,000 = 200 lbs

2. Find the probability that the weight of a randomly selected steer is between 1,739 and 1,800 lbs. Round your answer to four decimal places.

Let's find the z-score for 1,739 and 1.800, this way:

z-score = (X - μ)/σ

z-score = (1,739 - 1,400)/200

z-score = 339/200

z-score = 1.695 = 1.70 (rounding to the next hundredth)

z-score = (X - μ)/σ

z-score = (1,800 - 1,400)/200

z-score = 400/200

z-score = 2

Now, let's find p for z-score = 1.7 and p for z-score = 2, using the z-table, as follows:

p (z = 1.7) = 0.9554

p (z = 2) = 0.9772

In consequence,

p (1.7 ≤ z ≤ 2 ) = 0.9772 - 0.9554

p (1.7 ≤ z ≤ 2 ) = 0.0218

The probability that the weight of a randomly selected steer is between 1,739 and 1,800 lbs is 0.0218 or 2.18%

Answer 2

Answer:

$P(1739$

(Correct to 4 decimal places)

Step-by-step explanation:

The probability of a continuous normal variable X found in a particular interval [a, b] is the area under the curve bounded by x=a and x=b and is given by:

$P(a$

where

$f(X)=\frac{1}{\sigma \sqrt{2 \pi} }e^{-\frac{1}{2} (\frac{x-u}{ \sigma} )^2}$



$f(X)=\frac{1}{200 \sqrt{2 \pi} }e^{-\frac{1}{2} (\frac{x-1400}{ 200} )^2}\\=\frac{1}{200 \sqrt{2 \pi} }e^{- \frac{(x-1400)^2}{ 80000}$

$P(1739$

$P(1739$

Using a calculator,

$P(1739$