Answered: On one exam Mike had a score of 4,…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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On one exam Mike had a score of 4, which corresponded to a z-score of z = -1.50, and Sandy had a score of 11 that corresponded to a z-score of z = 2.00. Assuming the class has a normal distribution, what are the values for the mean and standard deviation for this exam?

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

The z-score formula is,

$z = \frac{x - μ}{σ}$

In the formula, x be the score, $μ$ be the population mean, and $σ$ be the population standard deviation.

On one exam Mike had a score of 4, which corresponded to a z-score of $z = - 1.50$ . That is,

$\begin{array}{rcl} - 1.50 & = & \frac{4 - μ}{σ} \\ - 1.50 σ & = & 4 - μ \\ μ - 1.50 σ - 4 & = & 0 . . . (1) \end{array}$

Sandy had a score of 11 that corresponded to a z-score of $z = 2.00$ . That is,

$\begin{array}{rcl} 2.00 & = & \frac{11 - μ}{σ} \\ 2 σ & = & 11 - μ \\ μ + 2.00 σ - 11 & = & 0 . . . (2) \end{array}$

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