Question

A half hour evening tv show has an average of 10.4 minutes of commercials. Assume the number of minutes of commercials follows a normal distribution with standard deviation of 2.0 minutes.

e) The lowest 25% of minutes of commercials is at most how long? (In other words, find the maximum minutes for the lowest 25% of number of minutes of commercials.) Use the nearest value in the z-table.

Answer 1

Answer: 9.06 minutes

Step-by-step explanation

In the back of your stats textbook is a bunch of tables. We'll need the Z table. If you don't have your textbook with you, then search out "z table" to find various alternatives.

Inside the table is a sea of values to wade through here. What we're looking for is 0.25 or the value closest to it.

Unfortunately 0.25 itself is not found here.

The closest we can get to 0.25 is the value 0.25143; this value is found in the row that starts with "-0.6" and the column that has "0.07" at the top. It indicates that P(Z < -0.67) = 0.25143 approximately.

In other words, roughly 25.143% of the standard normal distribution is to the left of z = -0.67

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We'll use this z value, along with mu = 10.4 and sigma = 2.0, to compute the value of x.

z = (x- mu)/sigma

z*sigma = x-mu

x = z*sigma + mu

x = -0.67*2.0 + 10.4

x = 9.06 is the answer

The lowest 25% of minutes of commercials is at most 9.06 minutes long. This value is an approximation. It's an approximation for two reasons

1. We didn't land on 0.25 exactly, and had to go with the closest value of 0.25143
2. Each value in the table is an approximation.

It basically tells us that P(X < 9.06) = 0.25143 when mu = 10.4 and sigma = 2.0

To get a better approximation, you'll need to use a stats calculator.