NCEA Level 2Normal distributionAS 91267

Normal distribution: 7 exam questions solved step by step

What this lesson covers

  • A normal distribution is symmetric about its mean, and its spread is measured by the standard deviation.
  • A z-value counts how many standard deviations a value is from the mean.
Key idea

7 questions, one method

  • 1. Sketch the curve, shade the area.
  • 2. z = (x − μ) ÷ σ, with its sign.
  • 3. Normal probability, then round.
  • 4. Check it against the sketch.
Your turn

Question 1 · Achieved

  • Kiwifruit weights at a packhouse are
  • normal: mean 110 g, sd 9 g.
  • Find P(a kiwifruit is under 122 g).
Question 1 · Achieved
Worked example

Q1: P(X < 122)

  • Shaded left of 122.
  • z first, then the area.
  1. μ = 110, σ = 9given
  2. P(X < 122)
  3. z = (122 − 110) ÷ 9
  4. = 12 ÷ 9 = 1.3333above: positive
  5. P(X < 122) = 0.908789
  6. = 0.90884 d.p.
  7. 0.9088 > 0.5over half is shaded
Your turn

Question 2 · Achieved

  • Same kiwifruit: mean 110 g, sd 9 g.
  • Find P(a kiwifruit is over 100 g).
  • Careful: 100 g is below the mean.
Question 2 · Achieved
Worked example

Q2: P(X > 100)

  • Below the mean: z < 0.
  • Shaded right: most of it.
  1. μ = 110, σ = 9given
  2. P(X > 100)
  3. z = (100 − 110) ÷ 9
  4. = −10 ÷ 9 = −1.1111below: negative
  5. P(X < 100) = 0.1333wrong side
  6. P(X > 100) = 0.866740
  7. = 0.86674 d.p.
  8. 0.8667 > 0.5most of the curve
Your turn

Question 3 · Merit

  • An Auckland bus route: trip times are
  • normal, mean 42 min, sd 5 min.
  • Find P(a trip takes 35 to 50 min).
Question 3 · Merit
Worked example

Q3: P(35 < X < 50)

  • Two values: two z-values.
  • One below, one above.
  1. μ = 42, σ = 5given
  2. P(35 < X < 50)
  3. z = (35 − 42) ÷ 5 = −1.4
  4. z = (50 − 42) ÷ 5 = 1.6
  5. P(−1.4 < Z < 1.6)same area
  6. P(35 < X < 50) = 0.864444
  7. = 0.86444 d.p.
  8. about 86% of trips35 to 50 min
Your turn

Question 4 · Merit, in context

  • Wellington ferry crossings are normal,
  • mean 55 min, sd 4 min.
  • Of 400 crossings, how many are
  • expected to take over 61 min?
Question 4 · Merit, in context
Worked example

Q4: how many over 61 min?

  • Probability first.
  • Then n × probability.
  1. μ = 55, σ = 4given
  2. P(X > 61)
  3. z = (61 − 55) ÷ 4 = 1.5
  4. P(X > 61) = 0.066807
  5. expected = 400 × 0.066807
  6. = 26.72
  7. about 27 of 400 crossingscrossings are whole
Your turn

Question 5 · Merit, in context

  • Honey jars for a school fundraiser:
  • normal, mean 508 g, sd 5 g.
  • Rejected if under 500 g or over 520 g.
  • What proportion are rejected?
Question 5 · Merit, in context
Worked example

Q5: two tails

  • Not symmetric: no doubling.
  • Each tail, then add.
  1. μ = 508, σ = 5given
  2. P(X < 500) + P(X > 520)
  3. z = (500 − 508) ÷ 5 = −1.6
  4. z = (520 − 508) ÷ 5 = 2.4
  5. P(X < 500) = 0.054799
  6. P(X > 520) = 0.008198
  7. 0.054799 + 0.008198
  8. = 0.062997 ≈ 0.06304 d.p.
  9. about 6.3% of jars rejected
Your turn

Question 6 · Excellence

  • The bus route again: mean 42, sd 5 min.
  • A trip has already taken over 45 min.
  • Find P(it takes over 50 min).
Question 6 · Excellence
Worked example

Q6: set it up

  • "Given" shrinks the total.
  • Over 50 is over 45.
  1. μ = 42, σ = 5given
  2. P(X > 50 | X > 45)given over 45
  3. P(A | B) = P(A and B) ÷ P(B)
  4. A: X > 50, B: X > 45
  5. A and B: X > 50over 50 is over 45
  6. = P(X > 50) ÷ P(X > 45)
Worked example

Q6: the two areas

  • P(X > 45) is the new total.
  • Round only at the end.
  1. P(X > 50) ÷ P(X > 45)
  2. z = (50 − 42) ÷ 5 = 1.6
  3. P(X > 50) = 0.054799
  4. z = (45 − 42) ÷ 5 = 0.6
  5. P(X > 45) = 0.274253the reduced total
  6. 0.054799 ÷ 0.274253
  7. = 0.199812
  8. = 0.19984 d.p.
  9. about 1 in 5 of those tripsover 45 min
Your turn

Question 7 · Excellence

  • Girls' 100 m: mean 13.8 s, sd 0.5 s.
  • Boys' 100 m: mean 12.5 s, sd 0.4 s.
  • Aroha runs 13.1 s. Tama runs 11.9 s.
  • Whose time is more impressive?
Question 7 · Excellence
Worked example

Q7: compare with z

  • Different spreads.
  • Faster is below: z < 0.
  1. Aroha 0.7 s, Tama 0.6 sspreads differ
  2. Aroha: z = (13.1 − 13.8) ÷ 0.5
  3. = −0.7 ÷ 0.5 = −1.4
  4. Tama: z = (11.9 − 12.5) ÷ 0.4
  5. = −0.6 ÷ 0.4 = −1.5
  6. −1.5 < −1.4further below
  7. Tama is 1.5 sd under his mean
Worked example

Q7: back it with areas

  • Who is faster than them?
  • Smaller area: rarer time.
  1. girls: μ = 13.8, σ = 0.5
  2. P(X < 13.1) = 0.080757
  3. = 0.08084 d.p.
  4. boys: μ = 12.5, σ = 0.4
  5. P(X < 11.9) = 0.066807
  6. = 0.06684 d.p.
  7. 0.0668 < 0.0808fewer are faster
  8. Tama's time is more impressive
Summary

The method that solved all seven

  • Sketch, find z, find the area, check, say it in context.
StepHow
Sketchshade the area you want
z-value(x − μ) ÷ σ, sign and all
Probabilitycalculator, round at the end
Two tailsfind each one, then add
Given thatdivide by the reduced total
Compareuse z, not raw units

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