the mean grade in physics examination was 75 and the standard deviation was 5. Determine the grade of the student whose standard is -0.5
μ=75σ=5\mu=75 \\ \sigma=5μ=75σ=5
Standard score z = -0.5
z=x−μσ−0.5=x−755−2.5=x−75x=75−25=50z = \frac{x- \mu}{\sigma} \\ -0.5 = \frac{x -75}{5} \\ -2.5 = x -75 \\ x = 75-25 = 50z=σx−μ−0.5=5x−75−2.5=x−75x=75−25=50
Answer: 50
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