Question

The amount of time to complete a physical activity in a PE class is approximately normally normally distributed with a mean of 32.9 seconds and a standard deviation of 6.4 seconds.

A) What is the probability that a randomly chosen student completes the activity in less than 33.2 seconds?
B) What is the probability that a randomly chosen student completes the activity in more than 46.6 seconds?
C) What proportion of students take between 35.5 and 42.8 seconds to complete the activity?
D) 75% of all students finish the activity in less than____seconds.

Answer 1

The answer is below

Explanation:

Given that mean (μ) of 32.9 seconds and a standard deviation (σ) of 6.4 seconds.

The z score is used to measure by how many standard deviation the raw score is above or below the mean. It is given by:

`z=(x-\mu)/(\sigma)\\`

a) For x < 33.2 seconds

`z=(x-\mu)/(\sigma)\\\\z=(33.2-32.9)/(6.4) =0.05`

From the normal distribution table, the probability that a randomly chosen student completes the activity in less than 33.2 seconds = P(x < 33.2) = P(z < 0.05) = 0.5199 = 51.99%

b) For x > 46.6 seconds

`z=(x-\mu)/(\sigma)\\\\z=(46.6-32.9)/(6.4) =2.14`

From the normal distribution table, the probability that a randomly chosen student completes the activity in more than 46.6 seconds = P(x > 46.6) = P(z > 2.14) = 1 - P(z < 2.14) = 1 - 0.9927 = 0.0073 = 0.73%

c) For x = 35.5 seconds

`z=(x-\mu)/(\sigma)\\\\z=(35.5-32.9)/(6.4) =0.41`

For x = 42.8 seconds

`z=(x-\mu)/(\sigma)\\\\z=(42.8-32.9)/(6.4) =1.55`

From the normal distribution table, the proportion of students take between 35.5 and 42.8 seconds to complete the activity = P(35.5 < x < 42.8) = P(0.41< z< 1.55) = P(z < 1.55) - P(z < 0.41) = 0.9332 - 0.6591 = 0.2741 = 27.41%

d) A probability of 75% = 0.75 corresponds to a z score of 0.68

`z=(x-\mu)/(\sigma)\\\\0.68=(x-32.9)/(6.4) \\\\x-32.9=4.4\\x=4.4+32.9\\x=37.3`

75% of all students finish the activity in less than 37.3 seconds