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Check Your Progress · Q20

Q.A company conducted an IQ test for randomly select 50 employees. Volunteer A scored 74 out of the possible 120 points. If the average IQ test score was recorded as 62 and the standard deviation was 11. How well did volunteer A perform on the test compared to the other volunteers?

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Volunteer A’s score of 7474 gives Z≈1.09Z\approx1.09, i.e. 1.091.09 standard deviations above the mean, placing A above roughly 86%86\% of the group.

Z=x−μσZ=\dfrac{x-\mu}{\sigma}

where xx = individual score, μ\mu = mean, σ\sigma = standard deviation.

Steps

  1. Given x=74x=74, μ=62\mu=62, σ=11.\sigma=11.

  2. Substitute: Z=74−6211=1211.Z=\dfrac{74-62}{11}=\dfrac{12}{11}.

  3. Compute: Z=1.0909≈1.09.Z=1.0909\approx1.09.

  4. Interpret with the Z-table: P(Z<1.09)=0.8621.P(Z<1.09)=0.8621.

  5. So A performed better than about 86.21%86.21\% of the volunteers (top ≈13.8%\approx13.8\%). …

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