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Marrrta [24]
3 years ago
12

A high school student took two college entrance exams, scoring 1070 on the SAT and 25 on the ACT. Suppose that SAT scores have a

mean of 950 and a standard deviation of 155 while the ACT scores have a mean of 22 and a standard deviation of 4. Assuming the performance on both tests follows a normal distribution, determine which test the student did better on.
Mathematics
1 answer:
antoniya [11.8K]3 years ago
8 0

Answer:

Due to the higher z-score, he did better on the SAT.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Determine which test the student did better on.

He did better on whichever test he had the higher z-score.

SAT:

Scored 1070, so X = 1070

SAT scores have a mean of 950 and a standard deviation of 155. This means that \mu = 950, \sigma = 155.

Z = \frac{X - \mu}{\sigma}

Z = \frac{1070 - 950}{155}

Z = 0.77

ACT:

Scored 25, so X = 25

ACT scores have a mean of 22 and a standard deviation of 4. This means that \mu = 22, \sigma = 4

Z = \frac{X - \mu}{\sigma}

Z = \frac{25 - 22}{4}

Z = 0.75

Due to the higher z-score, he did better on the SAT.

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Line A is parallel to line D.

Line A is perpendicular to line C.

Line C is perpendicular to line D.

=====================================================

Explanation:

Let's use the slope formula to calculate the slope of the line through (-1,-17) and (3,11)

(x_1,y_1) = (-1,-17) \text{ and } (x_2,y_2)  = (3,11)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{11 - (-17)}{3 - (-1)}\\\\m = \frac{11 + 17}{3 + 1}\\\\m = \frac{28}{4}\\\\m = 7\\\\

The slope of line A is 7

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Now let's find the slope of line B.

(x_1,y_1) = (0,4) \text{ and } (x_2,y_2)  = (7,-5)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{-5 - 4}{7 - 0}\\\\m = -\frac{9}{7}\\\\

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Now onto line C.

(x_1,y_1) = (7,1) \text{ and } (x_2,y_2)  = (0,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 1}{0 - 7}\\\\m = \frac{1}{-7}\\\\m = -\frac{1}{7}\\\\

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Lastly we have line D.

(x_1,y_1) = (-1,-6) \text{ and } (x_2,y_2)  = (1,8)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{8 - (-6)}{1 - (-1)}\\\\m = \frac{8 + 6}{1 + 1}\\\\m = \frac{14}{2}\\\\m = 7\\\\

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Here's a summary of the slopes we found

\begin{array}{|c|c|} \cline{1-2}\text{Line} & \text{Slope}\\\cline{1-2}\text{A} & 7\\\cline{1-2}\text{B} & -9/7\\\cline{1-2}\text{C} & -1/7\\\cline{1-2}\text{D} & 7\\\cline{1-2}\end{array}

Recall that parallel lines have equal slopes, but different y intercepts. This fact makes Line A parallel to line D.

Lines A and C are perpendicular to one another, because the slopes 7 and -1/7 multiply to -1. In other words, -1/7 is the negative reciprocal of 7, and vice versa. These two lines form a 90 degree angle.

Lines C and D are perpendicular for the same reasoning as the previous paragraph.

Line B unfortunately is neither parallel nor perpendicular to any of the other lines mentioned.

You can use a graphing tool like Desmos or GeoGebra to verify these answers.

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