14.8k views
2 votes
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 above 800?

The probability is__?

User Nokturnal
by
5.6k points

1 Answer

3 votes

Answer:

The probability is 0.0015

Explanation:

We know that the average
\mu is:


\mu=500

The standard deviation
\sigma is:


\sigma=100

The Z-score is:


Z=(x-\mu)/(\sigma)

We seek to find


P(x>800)

The Z-score is:


Z=(x-\mu)/(\sigma)


Z=(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)=0.0015

User Anarno
by
6.9k points