Question

A study conducted by the Child Mind Institute found that children spend, on average, 5.5 minutes in outdoor

unstructured play per day with a standard deviation of 1.24 minutes per day. Let X= the amount of time (in minutes)
that a randomly selected child spends in outdoor unstructured play per day. The probability distribution of X is
approximately normal. Find P(X28). Describe this probability in words.

Answer 1

The probability of X being less than 28 minutes is approximately 100%.

Step-by-step explanation:

To find the probability, we need to calculate the Z-score and use the standard normal distribution table.

The Z-score formula is: Z = (X - mean) / standard deviation.

In this case, X is 28 minutes, the mean is 5.5 minutes, and the standard deviation is 1.24 minutes.

Plugging these values into the formula, we get Z = (28 - 5.5) / 1.24 = 17.5 / 1.24 = 14.11.

Using the standard normal distribution table, we can find the probability associated with a Z-score of 14.11, which will give us the probability of X being less than 28 minutes.

The probability is approximately 1.000, or 100%.