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?

Question

Mathematics

Sebastian Pollard

29 February, 20:10

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?

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Rashad Potts · Answer 1

Step-by-step 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