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The times for completing one circuit of a bicycle course are normally distributed with a mean of 68.0 minutes and a standard deviation of 4.9 minutes. An association wants to sponsor a race but will cut the bottom 25% of riders. In a trial run, what should be the

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Answer:

64.697

Step-by-step explanation:

Given a normal distribution :

Mean = 68 minutes

Standard deviation = 4.9 minutes

To cut the bottom 25% riders

Using the relation :

Zscore = ( score - mean) / standard deviation

Zscore for the bottom 25% corresponds to

- 0.674(using tables or calculator)

Hence,

-0.674 = (score - 68) / 4.9

-0.674 * 4.9 = score - 68

-3.3026 = score - 68

-3.3026 + 68 = score

Score = 64.6974

Hence, those whose time is below 64.697 will be cut

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