Question

A college-entrance exam is designed so that scores are normally distributed with a mean of 500 and a standard deviation of 100. Using the eight‐part symmetry of the area under a normal curve, what is the probability that a randomly chosen exam score is less than 200 or greater than 800?
The probability is__?

Answer 1

Answer:

The probability is  0.003

Step-by-step explanation:

We know that the average $\mu$ is:

$\mu=500$

The standard deviation $\sigma$ is:

$\sigma=100$

The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

We seek to find

$P(x800)$

For P(x>800) The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{800-500}{100}$

$Z=3$

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

$P(x>800)=0.15\%$

For P(x<200) The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{200-500}{100}$

$Z=-3$

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

$P(x$

Therefore

$P(x800)=P(x800)$

$P(x800)=0.0015 + 0.0015$

$P(x800)=0.003$