Answer:
The probability is 0.003
Step-by-step explanation:
We know that the average [tex]\mu[/tex] is:
[tex]\mu=500[/tex]
The standard deviation [tex]\sigma[/tex] is:
[tex]\sigma=100[/tex]
The Z-score is:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]
We seek to find
[tex]P(x<200\ or\ x>800)[/tex]
For P(x>800) The Z-score is:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]
[tex]Z=\frac{800-500}{100}[/tex]
[tex]Z=3[/tex]
The score of Z = 3 means that 800 is 3 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 3 deviations from the mean has percentage of 0.15%
So
[tex]P(x>800)=0.15\%[/tex]
For P(x<200) The Z-score is:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]
[tex]Z=\frac{200-500}{100}[/tex]
[tex]Z=-3[/tex]
The score of Z = -3 means that 200 is 3 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 3 deviations from the mean has percentage of 0.15%
So
[tex]P(x<200)=0.0015[/tex]
Therefore
[tex]P(x<200\ or\ x>800)=P(x<200) +P(x>800)[/tex]
[tex]P(x<200\ or\ x>800)=0.0015 + 0.0015[/tex]
[tex]P(x<200\ or\ x>800)=0.003[/tex]
