SOLUTION: Let x denote the time taken to run a road race. Suppose x is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner i

Question 1178774: Let x denote the time taken to run a road race. Suppose x is approximately normally
distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If
one runner is selected at random, what is the probability that this runner will
complete this road race in:
a. less than 160 minutes? [3 marks]
b. greater than 210 minutes? [5 marks]
c. 215 to 245 minutes? [7 minutes]
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step:
**Given Information:**
* X ~ N(190, 21²) (X is normally distributed with mean μ = 190 minutes and standard deviation σ = 21 minutes)
**a. Less than 160 minutes?**
1. **Calculate the z-score:**
* z = (X - μ) / σ
* z = (160 - 190) / 21
* z = -30 / 21
* z ≈ -1.4286
2. **Find the probability:**
* P(X < 160) = P(z < -1.4286)
* Using a z-table or calculator, P(z < -1.43) ≈ 0.0764
**b. Greater than 210 minutes?**
1. **Calculate the z-score:**
* z = (X - μ) / σ
* z = (210 - 190) / 21
* z = 20 / 21
* z ≈ 0.9524
2. **Find the probability:**
* P(X > 210) = P(z > 0.9524)
* Using a z-table or calculator, P(z > 0.95) ≈ 0.1711
**c. 215 to 245 minutes?**
1. **Calculate the z-scores:**
* z₁ = (215 - 190) / 21
* z₁ = 25 / 21
* z₁ ≈ 1.1905
* z₂ = (245 - 190) / 21
* z₂ = 55 / 21
* z₂ ≈ 2.6190
2. **Find the probability:**
* P(215 < X < 245) = P(1.1905 < z < 2.6190)
* Using a z-table or calculator:
* P(z < 2.6190) ≈ 0.9956
* P(z < 1.1905) ≈ 0.8830
* P(1.1905 < z < 2.6190) = P(z < 2.6190) - P(z < 1.1905) ≈ 0.9956 - 0.8830 ≈ 0.1126

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