An instructor has given his final exam,and scores are approximately normally distributed.Nancy scores a 90 and Ronald scores a 75.The characteristics of the distribution are as follows:
= 85,s² = 36.What is Nancy's T score (for a mean of 50 and a standard deviation of 10) ?

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Quizplus · Accepted Answer

To calculate Nancy's T score, first find her Z score using the formula Z = (X - μ) / σ, where X is Nancy's score, μ is the mean, and σ is the standard deviation. Here, X = 90, μ = 85, and σ = √36 = 6. So, Z = (90 - 85) / 6 = 5/6 ≈ 0.8333. Then, convert the Z score to a T score using the formula T = 50 + 10 * Z. Thus, T = 50 + 10 * 0.8333 = 58.3.

An Instructor Has Given His Final Exam,and Scores Are Approximately