Assume that scores from a college admission test are normall

Assume that scores from a college admission test are normally distributed with a mean of 1,500 and a standard deviation of 300. Let x represent the test score.

                                      

                First Write the probability statement, and then do the calculation. Use 4-decimal place accuracy.

What proportion of the scores are between 1,000 and 1,800?

What proportion of the scores are above 2,000?

The top 10% of the scores are above what value?

The bottom 20% of the scores are below what value?

The middle 10% of the scores are between what two values?

Solution

Let x represent the test score

mean(X) = 1500

sd(X) = 300

(a) P(1000 < X < 1800) = P( (1000-1500)/300 < Z < (1800-1500)/300 )

= P(-1.6667 < Z < 1)

= P(Z < 1) - P(Z < -1.6667)

= 0.8413 - 0.0478

= 0.7935

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P(X > 2000) = P(Z > (2000 - 1500)/300 )

= P(Z > 1.6667)

= 0.0478

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let y be the top 10% scores

=> P(X > y) = 0.1

From Standard normal table, corresponding Z = 1.282

=> (y - 1500)/300 = 1.282

=> y = 1884.6

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let y be the bottom 20% scores

=> P(X < y) = 0.2

From Standard normal table, corresponding Z = -0.8416

=> (y - 1500)/300 = -0.8416

=> y = 1247.52

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let y1 and y2 be the scores under which 20% of the scores lie

=> P(y1 < X < y2) = 0.2

From Standard normal table, Z = 0.2533

=> (y2 - 1500)/300 = 0.2533

=> y2 = 1575.99

=> (y1 - 1500)/300 = -0.2533

=> y1 = 1424.01

Therefore, 20% of the scores lie between 1424.01 and 1575.99