Question #152039
Based on a study conducted on residents in the United States, we found their salaries follow the normally distribution with a mean of $48600, and a standard deviation of $14000.
a) If we randomly select 50 residents from the United States, approximately how many of them will have salaries above $70000?
b) If we randomly select 10 residents from the United States, what is the probability that their average salary is between $45000 to $50000?
c) Knowing Jackie’s salary is 17% below the third quartile, what is her actual salary?
1
Expert's answer
2020-12-28T14:46:03-0500

a)

P(X>70000)=P(Z>700004860014000)=P(Z>1.53)=0.063P(X > 70000) = P(Z > \frac{70000-48600}{14000}) \\ = P(Z> 1.53) \\ = 0.063

Therefore Expected number of people with salaries above $70000:

=0.063×50=3.15= 0.063 \times 50 = 3.15

b)

P(45000<Xˉ<50000)=P(45000486001400010<Z<50000486001400010)=P(0.81<Z<0.32)=P(Z<0.32)P(Z<0.81)=0.62550.2090=0.4165P(45000< \bar{X} < 50000) = P(\frac{45000-48600}{\frac{14000}{\sqrt{10}}} <Z< \frac{50000-48600}{\frac{14000}{\sqrt{10}}}) \\ = P(-0.81 < Z < 0.32) \\ = P(Z<0.32)-P(Z <-0.81) \\ = 0.6255 -0.2090 \\ = 0.4165

c) From standard normal tables:

P(Z<0.6745)=0.75P(Z < 0.6745) = 0.75

The third quartile salary:

=Mean+0.6745×σ=48600+0.6745×14000=58043= Mean + 0.6745 \times σ \\ = 48600 + 0.6745 \times 14000 \\ = 58043

Jackie's salary =0.83×58043=48175.69= 0.83 \times 58043 = 48175.69


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