Respuesta :
Answer:
The minimum score required for an A grade is 89.8.
Step-by-step explanation:
We are given that a psychology professor assigns letter grades on a test according to the following scheme. A : Top 7% of scores. B : Scores below the top 7% and above the bottom 64%. C : Scores below the top 36% and above the bottom 25%. D : Scores below the top 75% and above the bottom 6%. F : Bottom 6% of scores.
Scores on the test are normally distributed with a mean of 78.4 and a standard deviation of 7.6.
Let X = Scores on the test
SO, X ~ Normal([tex]\mu=78.4,\sigma^{2} =7.6^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean time = 78.4
[tex]\sigma[/tex] = standard deviation = 7.6
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, the minimum score required for an A grade so that it represents Top 7% of scores is given by;
P(X [tex]\geq[/tex] x) = 0.07 {where x is the required minimum score
P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{x-78.4}{7.6}[/tex] ) = 0.07
P(Z [tex]\geq[/tex] [tex]\frac{x-78.4}{7.6}[/tex] ) = 0.07
So, the critical value of x in the z table which represents the top 7% of the area is given as 1.4996, that is;
[tex]\frac{x-78.4}{7.6} =1.4996[/tex]
[tex]{x-78.4}{} =1.4996\times 7.6[/tex]
[tex]x[/tex] = 78.4 + 11.39696 = 89.8 or 90
Hence, the minimum score required for an A grade is 89.8.
