Answer:
The probability that he weighs between 170 and 220 pounds
P( 170 ≤ x≤ 200) = 0.3811
Step-by-step explanation:
<u><em>Explanation:</em></u>-
Given mean of the Population = 200 pounds
Given standard deviation of the Population = 50 pounds
Let 'X' be the random variable of the weights of college football players are normally distributed.
Let x₁ = 170


Let x₂ = 220


<em>The probability that he weighs between 170 and 220 pounds.</em>
<em>P( 170 ≤ x≤ 200) = P(Z₁ ≤z≤Z₂)</em>
= P(-0.6 ≤z≤0.4)
= P(z≤0.4)-P(z≤-0.6)
= 0.5 + A(0.4) -(0.5 - A(-0.6)
= A(0.4) + A(0.6) A(-z) = A(z)
= 0.1554 +0.2257 ( check<em> Areas in normal table)</em>
<em> = 0.3811</em>
<u><em>Conclusion</em></u><em>:-</em>
<em>The probability that he weighs between 170 and 220 pounds</em>
<em>P( 170 ≤ x≤ 200) = 0.3811</em>