The amount of time to complete a physical activity in a PE class is approximately normally normally distributed with a mean of 37.3 seconds and a standard deviation of 5.6 seconds.

Question

The amount of time to complete a physical activity in a PE class is approximately normally normally distributed with a mean of 37.3 seconds and a standard deviation of 5.6 seconds.

 

 What is the probability that a randomly chosen student completes the activity in less than 33.5 seconds? 

What is the probability that a randomly chosen student completes the activity in more than 43.8 seconds?

 What proportion of students take between 34.7 and 39.6 seconds to complete the activity?

85% of all students finish the activity in less than------- sec?

Answer 1

Normal distribution ("bell curve") with mean = 37.3 seconds and Standard Deviation = 5.6 seconds. For a bell curve, we should remember that, very approximately, 34% lies between mean & 1 SD, with further 14% between 1 & 2 SD (hence most of curve - 96% - lies within 2 SD of mean).

Z-score = (given value - mean) / SD

Negative z-score is left of mean, positive is right of mean. Convert z-score to percentage by looking up in Z-score table.

(1) For P(t < 33.5 seconds)

This is a bit less than 1 SD (=34%) to left of mean, so my guesstimate is 20% (50%-30%). Exact value requires us to calculate the z-score, as follows.

z = (33.5 - 37.3) / 5.6 = -0.68

from z-score table = 0.2483

hence P(t < 33.5 secs) = 24.8%

(2) For P(t > 43.8 seconds)

This is a bit more than 1 SD (34%) to right of mean, so my guesstimate is 10% (50% - 40%).

z = (43.8 - 37.3) / 5.6 = 1.16

from z-score table = 0.1230 (Note: here I looked up -1.16 z-score, using symmetry of bell curve.)

hence P(t < 43.8 secs) = 12.3%

(3) For P(34.7 secs < t < 39.6 secs)

34.7 is about 1/2 SD (left of mean), so 33% (50% - 1/2 of 34%). While 39.6 is also about 1/2 SD (but right of mean), so another 33%. Hence guesstimate of 34% (100-33-33).

z₁ = (34.7 - 37.3) / 5.6 = -0.464

from z-score table = 0.3210 = 32.1%

z₂ = (39.6 - 37.3) / 5.6 = 0.411

from z-score table = 0.3409 = 34.1% (again, I looked up z-score of -0.411.)

ALL - LESS THAN 34.7 - MORE THAN 39.6 = 100% - 32.1% - 34.1% = 33.8%

Hence P(34.7 secs < t < 39.6 secs) = 33.8%

(4) For P(t < x seconds) = 85%

For my guesstimate, mean is 50% and 1 SD is 34% so this is 84%. 1 SD is 42.9 secs.

(100-85) = 15% = 0.15 z-score table

=> z-score = 1.035

1.035 = (x - 37.3) / 5.6

=> x = 1.035 (5.6) + 37.3 = 43.1 seconds

Hence P(t < 43.1 secs) = 85%

Answer 2

Hi Zane,

Part A: 0.2487

Part B: 0.1229

Part C: 0.3381

Part D: 43.10 seconds

I would be happy to explain how I obtained these answers by scheduling a one-on-one tutoring session. I am using my TI-84 calculator and using the normalcdf and invNorm commands. You could also calculate these by hands by converting to z-scores and by using a z-table for an approximate answer.