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
[tex]Z = \frac{x-mean}{S.D} = \frac{920-1100}{300} = - 0.6[/tex]
Let x₂ = 1730
[tex]Z = \frac{x-mean}{S.D} = \frac{1730-1100}{300} = 2.1[/tex]
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