Let X= the weekly income: X∼N(μ,σ2)
Given that μ=800,σ=85
Z=σX−μ∼N(0,1)(i) The probability of finding a middle manager with a weekly income of between K840 and K900 is
P(840<X<900)=P(X<900)−P(X≤840)==P(Z<85900−800)−P(X≤85840−800)≈≈P(Z<1.176471)−P(X≤0.470588)≈≈0.88030−0.68103≈0.1993 (ii) The percent of middle managers that earn more than K905 is
P(X>905)=1−P(X≤905)==1−P(Z≤85905−800)≈1−P(Z≤1.23529)≈≈1−0.89164≈0.1084≈10.84% (iii) The percent of middle managers that earn less than K905 is
P(X<905)=P(Z<1.23529)≈≈0.89164≈89.16% (iv) The probability of finding a middle manager with weekly income of between K750 and K850 is
P(750<X<850)=P(X<850)−P(X≤750)==P(Z<85850−800)−P(X≤85750−800)≈≈P(Z<0.588235)−P(X≤−0.588235)≈≈0.72181−0.27819≈0.4436≈44.36% (v) The probability of finding a middle manager with a weekly income of between K700 and K790 is
P(700<X<790)=P(X<790)−P(X≤700)==P(Z<85790−800)−P(X≤85700−800)≈≈P(Z<−0.117647)−P(X≤−1.176471)≈≈0.45317−0.11970≈0.3335
(vi) Above what income would the top 10% of the managers earn?
P(Z>z∗)=0.1
z∗=85x∗−800≈1.28155
x∗≈909 (vii) Below what income would the lowest 10% of the managers earn?
P(Z<z∗)=0.1
z∗=85x∗−800≈−1.28155
x∗≈691
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