Answer:
<em>The approximate percentage of women with platelet counts between 127.7 and 378.5 </em>
<em>P( 127.7 ≤x≤378.5) = 0.9544 or 95 percentage</em>
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
<em>Mean of the Population = 253.1</em>
<em>Given standard deviation of the Population = 62.7</em>
<em>Given sample size 'n' = 873</em>
<em>Let 'X' be the random variable in Normal distribution</em>
<em>Let x₁ = 127.7</em>
![Z_{1} = \frac{x_{1} -mean}{S.D} = \frac{127.7-253.1}{62.7} = -2](https://tex.z-dn.net/?f=Z_%7B1%7D%20%3D%20%5Cfrac%7Bx_%7B1%7D%20-mean%7D%7BS.D%7D%20%3D%20%5Cfrac%7B127.7-253.1%7D%7B62.7%7D%20%3D%20%20-2)
Let x₂ = 378.5
![Z_{2} = \frac{x_{2} -mean}{S.D} = \frac{378.5-253.1}{62.7} = 2](https://tex.z-dn.net/?f=Z_%7B2%7D%20%3D%20%5Cfrac%7Bx_%7B2%7D%20-mean%7D%7BS.D%7D%20%3D%20%5Cfrac%7B378.5-253.1%7D%7B62.7%7D%20%3D%20%202)
<u><em>Step(ii)</em></u>:-
<em>The probability of women with platelet counts between 127.7 and 378.5.</em>
<em>P( 127.7 ≤x≤378.5) = P( -2≤Z≤2)</em>
<em> =</em> P(<em>Z≤2) - P(Z≤-2)</em>
<em> = 0.5 +A(2) - ( 0.5 - A(-2))</em>
<em> = A(2) + A(2) (∵A(-2) =A(2)</em>
<em> = 2 × A(2)</em>
<em> = 2× 0.4772</em>
<em> = 0.9544</em>
<u><em>Conclusion</em></u><em>:-</em>
<em>The approximate percentage of women with platelet counts between 127.7 and 378.5 is 0.9544 or 95 percentage</em>