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A psychologist regularly runs rats through a maze and records the amount of time it takes them to finish. One hundred rats were run through a maze and they averaged 20 seconds with a standard deviation of 5 seconds. Assume that run-times follow a mound-shaped distribution. What can you say about the percentage of rats that took less than 10 seconds to finish?

User Aluriak
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Answer:

Explanation:

  • Given that μ = 20seconds and SD-standard deviation = 5seconds
  • To get the percentage of rats that took less than 10 seconds to finish is;
  • P( X < 10) = P( z < x -μ/SD)
  • = P( z< -2.00) = 0.028 = 2.28%
  • Hence, 2.28% of the rats took less than 10seconds to finish
User Guerschon
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