The Normal Distribution and z-Scores — Solutions

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1. Fluency Show Solution

   85 = 100 − 15 = μ − σ  and  115 = 100 + 15 = μ + σ

   The range [μ − σ, μ + σ] contains 68% of all values by the 68–95–99.7% rule.

2. Fluency Show Solution

   177 = 165 + 12 = μ + 2σ

   95% of women have heights within μ ± 2σ, so 5% are outside this range. By symmetry, 2.5% are taller than 177 cm.

3. Fluency Show Solution

   z = (x − μ) / σ = (78 − 70) / 8 = 8/8 = 1

   A score of 78 is exactly 1 standard deviation above the mean.

4. Fluency Show Solution

   Rearrange: x = μ + zσ = 50 + (−1.5)(10) = 50 − 15 = 35

5. Understanding Show Solution

   μ ± 2σ = 65 ± 2(10) = 65 ± 20

   The middle 95% of scores fall between 45 and 85.

6. Understanding Show Solution

   Biology z-score: z = (82 − 75) / 7 = 7/7 = 1.0

   Chemistry z-score: z = (76 − 70) / 8 = 6/8 = 0.75

   Biology has the higher z-score (1.0 > 0.75), so Biology was the better relative performance. Maya was 1 full standard deviation above the Biology class mean, but only 0.75 standard deviations above the Chemistry class mean.

7. Understanding Show Solution

   (a) z = (38.0 − 36.8) / 0.4 = 1.2 / 0.4 = 3

   (b) The 99.7% rule states that 99.7% of values lie within μ ± 3σ. A temperature of 38.0°C falls exactly at μ + 3σ, meaning it is at the extreme boundary of the normal range. Only 0.3% ÷ 2 = 0.15% of readings would naturally be this high or higher. Yes, this is unusually high and clinically indicates a fever requiring attention.

8. Understanding Show Solution

   (a) Above 250: 250 = 200 + 2(25) = μ + 2σ. Outside μ ± 2σ = 5%; above μ + 2σ: 2.5%

   (b) Below 150: 150 = 200 − 2(25) = μ − 2σ. Below μ − 2σ: 2.5%

   (c) Between 175 and 225: 175 = μ − σ and 225 = μ + σ. Within μ ± σ: 68%

9. Problem Solving Show Solution

   (a) 484 = 500 − 2(8) = μ − 2σ and 516 = 500 + 2(8) = μ + 2σ.
   Within μ ± 2σ: 95% of boxes.

   (b) New μ = 510. We want: below 494 = 510 − 2(8) = μ − 2σ.
   Below μ − 2σ: (100% − 95%) / 2 = 2.5% of boxes.

   (c) Under the original setting (μ = 500g), 5% of boxes are outside μ ± 2σ. Under the recalibrated setting (μ = 510g), only 2.5% are underfilled (below 494g) while most boxes are well-filled. The recalibrated setting (510g) reduces underfilling. However, the original μ = 500g setting is centred on the label weight — context determines which is “better” (regulatory requirements vs. wastage).

10. Problem Solving Show Solution

    (a) Minimum mark = μ + zσ = 68 + 0.5(12) = 68 + 6 = 74%

    (b) Jordan’s z-score = (74 − 68) / 12 = 6/12 = 0.5.
    Jordan’s z-score exactly equals the threshold of 0.5, so Jordan is admitted (meets the minimum requirement).

    (c) 50% of students score above the mean (z > 0). Of the 50% between the mean and μ+σ, we need the portion above z = 0.5 (halfway to z = 1). Roughly 50% − 34%/2 ≈ 50% − 17% = 33% are above z = 0.5. More precisely, about 30–31% of students qualify. (For context, z = 0.5 corresponds to approximately the 69th percentile, meaning about 31% of students are admitted.)