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Answer:
The probability that the weight of a randomly selected steer is between 920 and 1730 lbs
<em> P(920≤ x≤1730) = 0.7078 </em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
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
Let x₂ = 1730
<u><em>Step(ii)</em></u>
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
<u><em>Conclusion:-</em></u>
The probability that the weight of a randomly selected steer is between 920 and 1730 lbs
<em> P(920≤ x≤1730) = 0.7078 </em>
1 . Consider lineups without x. You have to select
- 2 guards from 5 quards;
- 2 forwards from 3 forwards;
- 1 center from 4 centers.
It can be made in
different ways.
2. Consider lineups with x as guard. You have to select
- 2 guards from 6 quards (one of them must be x);
- 2 forwards from 3 forwards;
- 1 center from 4 centers.
It can be made in
different ways.
3. Consider lineups with x as forward. You have to select
- 2 guards from 5 quards;
- 2 forwards from 4 forwards (one of them must be x);
- 1 center from 4 centers.
It can be made in
different ways.
Therefore, the total number of different lineups is
120+60+120=300.