Question

Suppose that on a final exam in statistics the mean was 50 and the standard deviation was 10 . Find the following: (i) the standardized (z) scores of students receiving the following grades: $50,25,0,100,64$. (ii) the raw grades corresponding to standardized scores of $$ -2, \quad 2, \quad 1.95, \quad-2.58, \quad 1.65, \quad .33 . $$ 

   Suppose that on a final exam in statistics the mean was 50 and the standard deviation was 10 . Find the following:
(i) the standardized (z) scores of students receiving the following grades: $50,25,0,100,64$.
(ii) the raw grades corresponding to standardized scores of
$$
-2, \quad 2, \quad 1.95, \quad-2.58, \quad 1.65, \quad .33 .
$$
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Statistics; Probability, Inference, and Decision
Statistics; Probability, Inference, and Decision
William Lee Hays;… 1st Edition
Chapter 3, Problem 46 ↓

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The formula for calculating the z-score is: \[ z = \frac{X - \mu}{\sigma} \] where \( X \) is the raw score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.  Show more…

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Suppose that on a final exam in statistics the mean was 50 and the standard deviation was 10 . Find the following: (i) the standardized (z) scores of students receiving the following grades: $50,25,0,100,64$. (ii) the raw grades corresponding to standardized scores of $$ -2, \quad 2, \quad 1.95, \quad-2.58, \quad 1.65, \quad .33 . $$
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Key Concepts

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Standardization (z-scores)
Standardization is the process of converting raw scores into a common scale by subtracting the mean from the raw score and then dividing by the standard deviation. This transformation enables comparison across different datasets or distributions by expressing scores in terms of how many standard deviations they fall above or below the mean.
Reverse Transformation (Converting z-scores to Raw Scores)
The reverse process of standardization involves converting a z-score back to its corresponding raw score. This is done using the formula x = z × (standard deviation) + (mean). It is useful in recovering the original measurement from a standardized score or in determining what raw score corresponds to a specific position within the distribution.
Mean and Standard Deviation
The mean and standard deviation are fundamental parameters in statistics for describing a distribution. The mean provides the central tendency or average of the data, while the standard deviation quantifies the variability or spread of the data around that mean. Together, they allow for understanding the overall distribution and for performing standardization.
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On the final examination, the mean was 72 and the standard deviation was 15. 1) Determine the standard scores (z-values) of students receiving the grade of 65. 2) Students receiving the grade of 85. 3) Probability that a student will score greater than 75. 4) Probability that a student will score lower than 65. 5) Find the corresponding standard score of -0.50 and the corresponding score of 1.
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A set of final examination grades in an introductory statistics course is normally distributed, with a mean of 78 and a standard deviation of 8. Complete parts (a) through (d). a. What is the probability that a student scored below 87 on this exam? The probability that a student scored below 87 is . (Round to four decimal places as needed.) b. What is the probability that a student scored between 70 and 96? The probability that a student scored between 70 and 96 is . (Round to four decimal places as needed.) c. The probability is 5% that a student taking the test scores higher than what grade? The probability is 5% that a student taking the test scores higher than . (Round to the nearest integer as needed.) d. If the professor grades on a curve (for example, the professor could give A's to the top 10% of the class, regardless of the score), is a student better off with a grade of 94 on the exam with a mean of 78 and a standard deviation of 8 or a grade of 64 on a different exam, where the mean is 61 and the standard deviation is 3? Show your answer statistically and explain. A student is with a grade of 94 on the exam with a mean of 78 and a standard deviation of 8 because the Z value for the grade of 94 is and the Z value for the grade of 64 on the different exam is . (Round to two decimal places as needed.)
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