Suppose ACT Reading scores are normally distributed with a mean of 21.4 and a standard deviation of 5.9. A university plans to award scholarships to students whose scores are in the top 9%. What is the minimum score required for the scholarship? Round your answer to the nearest tenth, if necessary.

29.2

Step-by-step explanation:

Mean = 21.4

Standard deviation = 5.9%

The minimum score required for the scholarship which is the scores of the top 9% is calculated using the Z - Score Formula.

The Z- score formula is given as:

z = x - μ /σ

Z score ( z) is determined by checking the z score percentile of the normal distribution

In the question we are told that it is the students who scores are in the top 9%

The top 9% is determined by finding the z score of the 91st percentile on the normal distribution

z score of the 91st percentile = 1.341

Using the formula

z = x - μ /σ

Where

z = z score of the 91st percentile = 1.341

μ = mean = 21.4

σ = Standard deviation = 5.9

1.341= x - 21.4 / 5.9

Cross multiply

1.341 × 5.9 = x - 21.4

7.7526 = x -21.4

x = 7.7526 + 21.4

x = 29.1526

The 91st percentile is at the score of 29.1526.

We were asked in the question to round up to the nearest tenth.

Approximately, = 29.2

The minimum score required for the scholarship to the nearest tenth is 29.2 .

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2