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Chapter 1: Problem 23

The mean duration of television commercials is 75 seconds with standard deviation 20 seconds. Assume that the durations are approximately normally distributed to answer the following. a. What percentage of commercials last longer than 95 seconds? b. What percentage of the commercials last between 35 and 115 seconds? c. Would you expect commercial to last longer than 2 minutes? Why or why not?

Short Answer

Expert verified
a) 15.87% last longer than 95 seconds. b) 95.44% last between 35 and 115 seconds. c) It's unlikely a commercial lasts longer than 2 minutes (1.22% chance).

Step by step solution

01

Understanding the Problem

We are given that the mean duration of television commercials is 75 seconds and the standard deviation is 20 seconds. The distribution is approximately normal. We need to find percentages for certain durations.
02

Calculate Z-score for 95 seconds

To find the percentage of commercials lasting longer than 95 seconds, we need to calculate the Z-score. The Z-score is given by \(Z = \frac{X - \mu}{\sigma}\), where \(X = 95\), \(\mu = 75\), and \(\sigma = 20\). Therefore, \(Z = \frac{95 - 75}{20} = 1.0\).
03

Use the Z-table for 95 seconds

Using the standard normal distribution table (Z-table), we find the probability that a Z-score is less than 1.0. The Z-table gives \(P(Z < 1.0) = 0.8413\). Thus, \(P(Z > 1.0) = 1 - 0.8413 = 0.1587\). This means approximately 15.87% of commercials last longer than 95 seconds.
04

Calculate Z-scores for 35 and 115 seconds

Next, calculate the Z-scores for both 35 and 115 seconds. For 35 seconds, \(Z = \frac{35 - 75}{20} = -2.0\). For 115 seconds, \(Z = \frac{115 - 75}{20} = 2.0\).
05

Use the Z-table for 35 and 115 seconds

Find the probabilities using the Z-table. For \(Z = -2.0\), \(P(Z < -2.0) = 0.0228\). For \(Z = 2.0\), \(P(Z < 2.0) = 0.9772\). The probability that a commercial lasts between 35 and 115 seconds is \(P(-2.0 < Z < 2.0) = 0.9772 - 0.0228 = 0.9544\), or 95.44%.
06

Analyze commercials lasting more than 2 minutes

Two minutes is 120 seconds. Calculate the Z-score: \(Z = \frac{120 - 75}{20} = 2.25\). Using the Z-table, \(P(Z > 2.25) = 1 - P(Z < 2.25) = 0.0122\). Hence, only 1.22% of commercials are expected to last longer than 120 seconds, which makes it unlikely.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-score calculation
The Z-score is a crucial concept when working with normally distributed data. It helps us understand how far away a specific data point is from the mean. In essence, the Z-score tells us the number of standard deviations a data point is from the mean. To calculate the Z-score, use the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here:
  • \( X \) is the data point value.
  • \( \mu \) is the mean of the data set.
  • \( \sigma \) is the standard deviation of the data set.
Once you compute the Z-score, it provides a standardized way to assess the position of the data point within the distribution. For example, in the case of a television commercial lasting 95 seconds, the Z-score is 1.0, indicating that 95 seconds is one standard deviation above the mean of 75 seconds.
Standard normal distribution table
A Z-table, or standard normal distribution table, is an indispensable tool for finding probabilities related to a standard normal distribution. The Z-table provides cumulative probabilities from the mean up to a given Z-score.
When using the Z-table:
  • Find the Z-score you calculated in the step above.
  • Look up this Z-score in the table to find the probability that a score is less than this value.
For a Z-score of 1.0, the Z-table shows that approximately 84.13% of values are below it. Hence, we calculate probabilities for the complementary event by subtracting from 1. For example, if 84.13% are below a Z of 1.0, then:
  • 15.87% (i.e., 100% - 84.13%) are above it.
Using the table accurately provides insights into the percentage distribution above or between certain Z-scores.
Probability percentages
Probability percentages give us a way to say, "how likely is it?" for a particular outcome in a normal distribution scenario.
When computing probability percentages:
  • Calculate the Z-scores of your desired data points.
  • Use these Z-scores to determine probabilities using the Z-table.
  • Translate these probabilities into percentages by multiplying by 100.
For instance, when considering a commercial lasting between 35 and 115 seconds, Z-scores provide the boundaries. The calculation reveals a 95.44% probability that commercials fit within those bounds.
Similarly, a 1.22% probability for commercials exceeding 120 seconds indicates rarity. Hence, the percentages derived from these probabilities help in making predictions and understanding distribution characteristics effectively.

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Most popular questions from this chapter

Of great importance to residents of central Florida is the amount of radioactive material present in the soil of reclaimed phosphate mining areas. Measurements of the amount of \(^{238} \mathrm{U}\) in 25 soil samples were as follows (measurements in picocuriespergram):$$\begin{array}{rrrrr}.74 & 6.47 & 1.90 & 2.69 & .75 \\\\.32 & 9.99 & 1.77 & 2.41 & 1.96 \\\1.66 & .70 & 2.42 & .54 &3.36 \\\3.59 & .37 & 1.09 & 8.32 & 4.06 \\ 4.55 & .76 & 2.03 & 5.70 & 12.48\end{array}$$.Construct a relative frequency histogram for these data.

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