A professor at a local university noted that the grades of her students were normally distributed with a mean of 78 and a standard deviation of 10. The professor has informed us that 16.6 percent of her students received grades of A. What is the minimum score needed to receive a grade of A?

We are given that a professor at a local university noted that the grades of her students were normally distributed with a mean of 78 and a standard deviation of 10.

The professor has informed us that 16.6 percent of her students received grades of A.

Let X = grades of the students

SO, X ~ Normal([tex]\mu=78,\sigma^{2} =10^{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 grades = 78

[tex]\sigma[/tex] = standard deviation = 10

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, we are given that the professor has informed us that 16.6 percent of her students received grades of A, so the minimum score needed to receive grade A is given by;

P(X [tex]\geq[/tex] x) = 0.166 {where x is the required minimum score needed}

P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{x-78}{10}[/tex] ) = 0.166

P(Z [tex]\geq[/tex] [tex]\frac{x-78}{10}[/tex] ) = 0.166

So, the critical value of x in the z table which represents the top 16.6% of the area is given as 0.9741, that is;

[tex]\frac{x-78}{10} =0.9741[/tex]

[tex]{x-78}{} =0.9741\times 10[/tex]

[tex]x[/tex] = 78 + 9.741 = 87.74

Hence, the minimum score needed to receive a grade of A is 87.74.