Question #89650
5. Determine the z-score value in each of the following scenarios:
a. What z-score value separates the top 8% of a normal distribution from the bottom
92%?
b. What z-score value separates the top 72% of a normal distribution from the bottom
28%?
c. What z-score value form the boundaries for the middle 58% of a normal
distribution?
d. What z-score value separates the middle 45% from the rest of the distribution?
2. Annual salaries for a large company are approximately normally distributed with a mean of
R50,000 and a standard deviation of R20,000.
a. What salary would an employee need to get in order to be in the lowest 30%?
b. What is the probability of having an above average salary range of between R60000 to
R80000.
1
Expert's answer
2019-05-14T10:25:22-0400

a.

P(Z<z)=0.92P(Z<z^*)=0.92z=1.4051z^*=1.4051

b.


P(Z<z)=0.28P(Z<z^*)=0.28

z=0.5828z^*=-0.5828

c.


P(z1<Z<z2)=0.58P(z^*_1<Z<z^*_2)=0.58

0.50.58/2=0.210.5-0.58/2=0.21

P(Z<z1)=0.21P(Z<z^*_1)=0.21

z1=0.8064z^*_1=-0.8064

z=±0.8064z^*=\pm 0.8064


So the interval is (0.8064,0.8064)(-0.8064, 0.8064)  


d.


0.50.45/2=0.2750.5-0.45/2=0.275

P(Z<z1)=0.275P(Z<z^*_1)=0.275

z1=0.5978z^*_1=-0.5978

z=±0.5978z^*=\pm 0.5978

So the interval is (0.5978,0.5978).(-0.5978, 0.5978).


2.


μ=R50,000\mu=R50,000σ=R20,000\sigma=R20,000

a. What salary would an employee need to get in order to be in the lowest 30%? 

Z=Xμσ=XμσZ={X-\mu \over \sigma}={X-\mu \over \sigma}

P(Z<z)=0.3=>z=0.5244P(Z<z^*)=0.3=>z^*=-0.5244

x=zσ+μx=z^*\sigma+\mu

x=0.5244(R20000)+R50000=R39512x=-0.5244(R20000)+R50000=R39512

b. What is the probability of having an above average salary range of between R60000 to

R80000.


z1=R60000R50000R20000=0.5z_1={R60000-R50000 \over R20000}=0.5

z2=R80000R50000R20000=1.5z_2={R80000-R50000 \over R20000}=1.5

P(R60000<X<R80000)=P(0.5<Z<1.5)=P(R60000<X<R80000)=P(0.5<Z<1.5)=

=P(Z<1.5)P(Z<0.5)==P(Z<1.5)-P(Z<0.5)=

=0.933192800.69146246=0.24173034=0.93319280-0.69146246=0.24173034

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