Question

The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 1100lbs. Find the probability that the weight of a randomly selected steer is between 920 and 1730lbs round to four decimal places.

Answer 1

Answer:

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(920≤ x≤1730) = 0.7078

Step-by-step explanation:

Step(i):-

Given mean of the Population = 1100 lbs

Standard deviation of the Population = 300 lbs

Let 'X' be the random variable in Normal distribution

Let x₁ = 920

$Z = \frac{x-mean}{S.D} = \frac{920-1100}{300} = - 0.6$

Let x₂ = 1730

$Z = \frac{x-mean}{S.D} = \frac{1730-1100}{300} = 2.1$

Step(ii)

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(x₁≤ x≤x₂) = P(Z₁≤ Z≤ Z₂)

                  = P(-0.6 ≤Z≤2.1)

                  = P(Z≤2.1) - P(Z≤-0.6)

                 = 0.5 + A(2.1) - (0.5 - A(-0.6)

                 =  A(2.1) +A(0.6)               (∵A(-0.6) = A(0.6)

                 =  0.4821 + 0.2257

                 = 0.7078

Conclusion:-

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

            P(920≤ x≤1730) = 0.7078