Question 4

The amount of time to complete a physical activity in a PE class is approximately normally distributed with a mean of 35.2 seconds and a standard deviation of 5.3 seconds.
a) What is the probability that a randomly chosen student completes the activity in less than 32.6 seconds?

Step 1 ：The problem is asking for the probability that a student completes a physical activity in less than 32.6 seconds. The time to complete the activity is normally distributed with a mean of 35.2 seconds and a standard deviation of 5.3 seconds.

Step 2 ：We can use the cumulative distribution function (CDF) of the normal distribution to find this probability. The CDF gives the probability that a random variable is less than or equal to a certain value.

Step 3 ：First, we calculate the z-score, which is the number of standard deviations a data point is from the mean. The formula for the z-score is \(z = \frac{X - \mu}{\sigma}\), where \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 4 ：Substituting the given values into the formula, we get \(z = \frac{32.6 - 35.2}{5.3} = -0.49\).

Step 5 ：The z-score tells us that 32.6 seconds is 0.49 standard deviations below the mean.

Step 6 ：Next, we use the standard normal distribution table or a calculator to find the probability associated with this z-score. The probability that a randomly chosen student completes the activity in less than 32.6 seconds is approximately 0.312.

Step 7 ：This means that about 31.2% of students are expected to complete the activity in less than 32.6 seconds.

Step 8 ：Final Answer: The probability that a randomly chosen student completes the activity in less than 32.6 seconds is approximately \(\boxed{0.312}\).