To solve this problem, we will convert the given values into z-scores and use the standard normal distribution table.
The z-score formula is given by:
z = (x - μ) / σ
where:
x = value we want to convert to a z-score (168 minutes),
μ = population mean (190 minutes),
σ = population standard deviation (21 minutes).
Substituting the given values:
z = (168 - 190) / 21
z = -22 / 21
z ≈ -1.048
Next, we can use the z-score to find the probability using the standard normal distribution table. However, since we want to find the probability of completing the race in less than 168 minutes, we need to find the area to the left of the z-score.
Using the standard normal distribution table or a calculator, we find that the probability associated with a z-score of -1.048 is approximately 0.1459.
Therefore, the probability that a runner will complete this road race in less than 168 minutes is approximately 0.1459, or 14.59%.
Let x denote the time it takes 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 less than 168 minutes?
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