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Normal Distribution - Find the Value of X

This ppt covered Normal Distribution - Find the Value of X using normal curve. Which helps in solving problem when Value of X is unknown

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Normal Distribution - Find the Value of X

  1. 1. Normal Distribution –Find the value of X SUNDARA B. N. Assistant Professor
  2. 2. Problem In a university the mean weight of 1000 male students is 60kg and Standard Deviation is 16kg a) Find the number of male students having their weights i. Less than 55kg ii. More than 70 kg iii. Between 45 kg and 65 kg b) What is the lowest weight of the 100 heaviest male students?
  3. 3. Given Information Here we are given that N=1000 μ = 60 σ = 16  X~N(60,256), We know that is X~N(μ, σ²), then the S.N.V is given by Z = X – μ σ
  4. 4. i) Less than 55kg Consider X = 55  P(X<55)=P(Z<- 0.31) =P(Z>0.31) = 0.5-P(0<Z<0.31) = 0.5-0.1217(TV) =0.3783 No. of male students having weight less than 55kg = NxP(X<55) = 1000x0.3783 =378 Z=-0.31 Z=0 X=6 0 X=5 5
  5. 5. ii) More than 70 kg Consider X = 70  P(X>70)=P(Z>0.63) = 0.5- P(0<Z<0.63) = 0.5-0.2357(TV) =0.2643 No. of male students having weight more than 70kg = NxP(X>70) = 1000x0.2643 =264 Z=0 Z=0.63 X=6 0 X=7 2
  6. 6. iii) Between 45 kg and 65 kg Consider X = 45 and X = 65  P(45<X<65)=P(-.94<X<0.31) = P(-0.94<X<0)+P(0<Z<0.31) = P(0<Z<0.94)+P(0<Z<0.31) = 0.3264+0.1217 (TV) =0.4481 No. of male students having weight between 45kg & 65kg = P(45<X<65) = 1000x0.4481 =448 X=8 0 Z=0 X=6 5 Z=0.3 1 X=4 5 Z=-0.94
  7. 7. b) What is the lowest weight of the 100 heaviest male students? Let x₁ be the lowest weight amongst 100 heaviest students. Now, for X=x₁, P(Z≥z₁)= P(0≤Z≤z₁) = 0.1 P(0≤Z≤z₁) = 0.5-0.1=0.4 z₁= 1.28 (From Table) x₁=60+16x1.28=60+20.48=80. 48  the lowest weight of 100 heaviest male students is 80.48kg X=μ X=x₁

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