Curriculum
Statistics — Third Secondary
Unit 1: Correlation and Regression
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Unit 2: Advanced Statistical Measures
0/2
Unit 3: Probability
0/2
Unit 4: Random Variables
0/4
Unit 5: The Normal Distribution
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Text lesson
The Normal Distribution and Its Applications
The Normal Distribution and Its Applications
Worked examples
Worked examples
Example 1→For a standard normal variable Z, find (1) P(0 ≤ Z ≤ 1), (2) P(Z ≥ 1), (3) P(−1 ≤ Z ≤ 1).
- P(0 ≤ Z ≤ 1) = 0.3413 from the table.
- P(Z ≥ 1) = 0.5 − 0.3413 = 0.1587.
- P(−1 ≤ Z ≤ 1) = 2 × 0.3413 = 0.6826.
- Example 2 The height of a type of plant follows a normal distribution with a mean of 40 cm and a standard deviation of 5 cm. What percentage of the plants are taller than 50 cm?
- X ~ N(40, 5²), and we need P(X ≥ 50).
- Z = (50 − 40) ÷ 5 = 2, so P(X ≥ 50) = P(Z ≥ 2).
- P(Z ≥ 2) = 0.5 − P(0 ≤ Z ≤ 2) = 0.5 − 0.4772 = 0.0228.
- The percentage = 0.0228 × 100% = about 2.28%.
- Example 3 For the same variable X ~ N(40, 5²), find the percentage of plants whose height lies between 35 cm and 45 cm.
- The Z-value for 35 = (35 − 40) ÷ 5 = −1, and for 45 = (45 − 40) ÷ 5 = 1.
- P(35 ≤ X ≤ 45) = P(−1 ≤ Z ≤ 1) = 2 × 0.3413 = 0.6826.
- The percentage is about 68.26%, which agrees with the rule that about 68% of values lie within one standard deviation of the mean.
- Example 4 (reversed problem) For the same variable X ~ N(40, 5²), find the height a such that 97.72% of the plants are shorter than a.
- We need P(X ≤ a) = 0.9772.
- Since P(Z ≤ z) = 0.5 + P(0 ≤ Z ≤ z), we get P(0 ≤ Z ≤ z) = 0.9772 − 0.5 = 0.4772.
- From the table z = 2, so (a − 40) ÷ 5 = 2.
- a − 40 = 10, so a = 50 cm.
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