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.

(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:

where



Using a calculator,

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%

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above