A large number of applicants for admission to graduate study in business are given an aptitude test. Scores are normally distributed with a mean of 460 and standard deviation of 80. What fraction of applicants would you expect to have scores of 600 or above?

4.01% of the applicants would be expected to have scores of 600 or above.

The z score is used to determine by how many standard deviations the raw score is above or below the mean.

It is given by:

Given that:

μ = 460, σ = 80. Hence for x > 600:

From the normal distribution table, P(x > 600) = P(z > 1.75) = 1 - P(z < 1.75) = 1 - 0.9599 = 0.0401 = 4.01%

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the fraction or percentage of the applicants that we would expect to have a score of 600 or above is 4.006%

Step-by-step explanation:

Scores are normally distributed with a mean of 460 and a standard deviation of 80.

Thus, for a value x, the associated z-score is given as;

z = (x - 460)/80

So, the Z-score for x = 600 is;

z = (600 - 460)/80

z = 1.75

From tbe tablw attached, The probability is;

P(Z > 1.75) = 1 - 0.95994

P(Z > 1.75) = 0.04006

Thus, the fraction or percentage of the applicants that we would expect to have a score of 600 or above is 4.006%

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step-by-step explanation:

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= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

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= sin (pi + pi/6) - sin pi/3  

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