Chapter 2: Q 35. (page 109)

Scores on the ACT college entrance exam follow a bell-shaped distribution with a mean $18$ and a standard deviation $6$ . Wayne’s standardized score on the ACT was $- 0.7$ What was Wayne’s actual ACT score? $(a) 4.2 (b) - 4.2 (c) 13.8 (d) 17.3 (e) 22.2$

Short Answer

Expert verified

The correct option is (c).

Step by step solution

Step 1. Given

The average population $(μ) = 18$

The standard deviation of the population $(σ) = 6$

A score that is standardized $(z) = - 0.7$

Step 2. Concept

The z score is calculated using the following formula: $z = X - μ / σ$

Step 3. Calculation

Consider that $X$ reflects Wayne's real ACT score, which follows a normal distribution with a mean of $18$ and a standard deviation of $6$ .

Wayne's real ACT score is determined as follows:

$z = \frac{X - μ}{σ}$

$- 0.7 = \frac{X - 18}{6}$

$- 4.2 + 18 = X$

$X = 13.8$

Hence, the correct option is $(c) 13.8$

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Scores on the ACT college entrance exam follow a bell-shaped distribution with a mean $18$ and a standard deviation $6$ . Wayne’s standardized score on the ACT was $- 0.7$ What was Wayne’s actual ACT score? $(a) 4.2 (b) - 4.2 (c) 13.8 (d) 17.3 (e) 22.2$

Short Answer

Step by step solution

Step 1. Given

Step 2. Concept

Step 3. Calculation

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Scores on the ACT college entrance exam follow a bell-shaped distribution with a mean 18 and a standard deviation 6. Wayne’s standardized score on the ACT was −0.7 What was Wayne’s actual ACT score? (a)4.2(b)−4.2(c)13.8(d)17.3(e)22.2

Short Answer

Step by step solution

Step 1. Given

Step 2. Concept

Step 3. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Mechanics Maths

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Study anywhere. Anytime. Across all devices.

Scores on the ACT college entrance exam follow a bell-shaped distribution with a mean $18$ and a standard deviation $6$ . Wayne’s standardized score on the ACT was $- 0.7$ What was Wayne’s actual ACT score? $(a) 4.2 (b) - 4.2 (c) 13.8 (d) 17.3 (e) 22.2$