Answer:
The answer is below
Step-by-step 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:
[tex]z=\frac{x-\mu}{\sigma}\\[/tex]
a) For x < 33.2 seconds
[tex]z=\frac{x-\mu}{\sigma}\\\\z=\frac{33.2-32.9}{6.4} =0.05[/tex]
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
[tex]z=\frac{x-\mu}{\sigma}\\\\z=\frac{46.6-32.9}{6.4} =2.14[/tex]
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
[tex]z=\frac{x-\mu}{\sigma}\\\\z=\frac{35.5-32.9}{6.4} =0.41[/tex]
For x = 42.8 seconds
[tex]z=\frac{x-\mu}{\sigma}\\\\z=\frac{42.8-32.9}{6.4} =1.55[/tex]
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
[tex]z=\frac{x-\mu}{\sigma}\\\\0.68=\frac{x-32.9}{6.4} \\\\x-32.9=4.4\\x=4.4+32.9\\x=37.3[/tex]
75% of all students finish the activity in less than 37.3 seconds
