Answer:
The approximate percentage of women with platelet counts between 127.7 and 378.5
P( 127.7 ≤x≤378.5) = 0.9544 or 95 percentage
Explanation:
Step(i):-
Mean of the Population = 253.1
Given standard deviation of the Population = 62.7
Given sample size 'n' = 873
Let 'X' be the random variable in Normal distribution
Let x₁ = 127.7
![Z_(1) = (x_(1) -mean)/(S.D) = (127.7-253.1)/(62.7) = -2](https://img.qammunity.org/2021/formulas/mathematics/college/v2z3m4jhru5cs964qnqbwkp03irqfptrmi.png)
Let x₂ = 378.5
![Z_(2) = (x_(2) -mean)/(S.D) = (378.5-253.1)/(62.7) = 2](https://img.qammunity.org/2021/formulas/mathematics/college/wo9qgglqo69ij9mz00kykh0q8wsmyvnyy4.png)
Step(ii):-
The probability of women with platelet counts between 127.7 and 378.5.
P( 127.7 ≤x≤378.5) = P( -2≤Z≤2)
= P(Z≤2) - P(Z≤-2)
= 0.5 +A(2) - ( 0.5 - A(-2))
= A(2) + A(2) (∵A(-2) =A(2)
= 2 × A(2)
= 2× 0.4772
= 0.9544
Conclusion:-
The approximate percentage of women with platelet counts between 127.7 and 378.5 is 0.9544 or 95 percentage