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In a distribution exactly normal 7% of items are under 35 and 89% of items are under 63. Find the probability that an item selected at random lies between 45 and 56.

Subject: Applied Mathematics 2

Topic: Probability Distribution

Difficulty: High

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Since 7% items are below 35 ,i.e. 50 – 7 = 43% items are between 35 & m & since 89% items are below 63 ,89 – 50 =39% items are between m ,and 63.

For area 0.43 , Z =1.48 , since 35 is less than m ,Z= -1.48 and for area 0.39 ,Z=1.23

∴ $\frac{(35-m)}{σ}$ = -1.48 & $\frac{(63-m)}{σ}$ = 1.23

solving above equations , we get m =50.3 & σ =10.33

Now Z = $\frac{(X-m)}{σ}$ = $\frac{(X-50.3)}{10.33}$

When X=45 , Z = $\frac{(45-50.3)}{10.33}$ = - 0.51

When X=56 , Z = $\frac{(56-50.3)}{10.33}$ = 0.55

∴ P( 45≤X≤ 56) = P( - 0.51 ≤ Z≤ 0.55) = Area from Z=0 to Z=0.51 + area from Z=0 to Z= 0.55

= 0.1950+ 0.2088 = 0.4038