a.
P(Z<z∗)=0.92z∗=1.4051 b.
P(Z<z∗)=0.28
z∗=−0.5828 c.
P(z1∗<Z<z2∗)=0.58
0.5−0.58/2=0.21
P(Z<z1∗)=0.21
z1∗=−0.8064
z∗=±0.8064
So the interval is (−0.8064,0.8064)
d.
0.5−0.45/2=0.275
P(Z<z1∗)=0.275
z1∗=−0.5978
z∗=±0.5978 So the interval is (−0.5978,0.5978).
2.
μ=R50,000σ=R20,000 a. What salary would an employee need to get in order to be in the lowest 30%?
Z=σX−μ=σX−μ
P(Z<z∗)=0.3=>z∗=−0.5244
x=z∗σ+μ
x=−0.5244(R20000)+R50000=R39512 b. What is the probability of having an above average salary range of between R60000 to
R80000.
z1=R20000R60000−R50000=0.5
z2=R20000R80000−R50000=1.5
P(R60000<X<R80000)=P(0.5<Z<1.5)=
=P(Z<1.5)−P(Z<0.5)=
=0.93319280−0.69146246=0.24173034
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