Answer:
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%
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:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]
Given that:
ΞΌ = 460, Ο = 80. Hence for x > 600:
[tex]z=\frac{600-460}{80} \\\\z=1.75[/tex]
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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