According to the Nielsen Company, the mean number of TV sets in a U.S. household in 2013 was 2.24. Assume the standard deviation is 1.4. A sample of 85 households is drawn.

Question

(a) What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to at least four decimal places.

Answer 1

To calculate the probability that the sample mean number of TV sets is greater than 2, we can use the Central Limit Theorem (CLT). According to CLT, the distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large.

Given:

• Population mean (\( \mu \)) = 2.24
• Population standard deviation (\( \sigma \)) = 1.4
• Sample size (\( n \)) = 85

Step 1: Calculate the standard error (SE) of the sample mean

The formula for the standard error of the sample mean is: \[ SE = \frac{\sigma}{\sqrt{n}} \] Substituting the given values: \[ SE = \frac{1.4}{\sqrt{85}} \approx \frac{1.4}{9.2195} \approx 0.151 \]

Step 2: Calculate the z-score for a sample mean of 2

To find the z-score, we use the formula: \[ z = \frac{\bar{x} - \mu}{SE} \] Where:

• \( \bar{x} \) = 2
• \( \mu \) = 2.24
• \( SE \) = 0.151

Substituting the values: \[ z = \frac{2 - 2.24}{0.151} \approx \frac{-0.24}{0.151} \approx -1.587 \]

Step 3: Find the probability for this z-score

Now we need to find the probability that a z-score is greater than -1.587. We can look this up in a standard normal distribution table or use a calculator.

Using a standard z-table or calculator:

• The cumulative probability for \( z = -1.587 \) is approximately 0.0571.

Since we want the probability that the sample mean is greater than 2, we calculate: \[ P(Z > -1.587) = 1 - P(Z < -1.587) = 1 - 0.0571 = 0.9429 \]

Final Answer

Thus, the probability that the sample mean number of TV sets is greater than 2 is approximately: \[ \boxed{0.9429} \]

Answer 2

(b) What is the probability that the sample mean number of TV sets is between 2.5 and 3? Round your answer to at least four decimal places.

Answer 3

To find the probability that the sample mean number of TV sets is between 2.5 and 3, we will again standardize these values using the z-score formula and then find the corresponding probabilities.

Given:

• Population mean (\( \mu \)) = 2.24
• Population standard deviation (\( \sigma \)) = 1.4
• Sample size (\( n \)) = 85

Step 1: Calculate the standard error (SE) of the sample mean

As calculated previously: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{85}} \approx 0.151 \]

Step 2: Calculate the z-scores for 2.5 and 3

Z-score for 2.5:

\[ z_1 = \frac{2.5 - 2.24}{0.151} \approx \frac{0.26}{0.151} \approx 1.719 \]

Z-score for 3:

\[ z_2 = \frac{3 - 2.24}{0.151} \approx \frac{0.76}{0.151} \approx 5.034 \]

Step 3: Find the corresponding probabilities

Using a standard normal distribution table or calculator:

1. For \( z_1 = 1.719 \):

   • \( P(Z < 1.719) \approx 0.9573 \) (Using a z-table or calculator)

2. For \( z_2 = 5.034 \):

   • \( P(Z < 5.034) \approx 1 \) (Since this is far out in the tail of the normal distribution)

Step 4: Calculate the probability that the sample mean is between 2.5 and 3

The probability that the sample mean is between 2.5 and 3 is: \[ P(2.5 < \bar{X} < 3) = P(Z < z_2) - P(Z < z_1) \approx 1 - 0.9573 = 0.0427 \]

Final Answer

Thus, the probability that the sample mean number of TV sets is between 2.5 and 3 is approximately: \[ \boxed{0.0427} \]

Answer 4

(c) Find the 90th percentile of the sample mean. Round the answer to at least two decimal places.

Answer 5

To find the 90th percentile of the sample mean, we will use the properties of the normal distribution. First, we will determine the z-score that corresponds to the 90th percentile, and then we will convert this z-score back to the sample mean using the mean and the standard error.

Given:

• Population mean (\( \mu \)) = 2.24
• Population standard deviation (\( \sigma \)) = 1.4
• Sample size (\( n \)) = 85

Step 1: Calculate the standard error (SE)

As previously calculated: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{85}} \approx 0.151 \]

Step 2: Find the z-score for the 90th percentile

From the standard normal distribution, the z-score that corresponds to the 90th percentile is approximately \( z \approx 1.2816 \).

Step 3: Convert the z-score to the sample mean

We can use the following formula to find the sample mean at the 90th percentile: \[ \bar{x} = \mu + z \cdot SE \] Substituting the values: \[ \bar{x} = 2.24 + 1.2816 \cdot 0.151 \] Calculating the second term: \[ 1.2816 \cdot 0.151 \approx 0.1934 \] Now, adding this to the mean: \[ \bar{x} = 2.24 + 0.1934 \approx 2.4334 \]

Final Answer

Rounding to two decimal places, the 90th percentile of the sample mean number of TV sets is approximately: \[ \boxed{2.43} \]

Answer 6

(d) Would it be unusual for the sample mean to be less than 2? Round your answer to at least four decimal places.

why or why not

Answer 7

To determine whether it would be unusual for the sample mean number of TV sets to be less than 2, we can calculate the z-score for a sample mean of 2 and then find the corresponding probability. If this probability is low (commonly below 0.05), it may be considered unusual.

Given:

• Population mean (\( \mu \)) = 2.24
• Population standard deviation (\( \sigma \)) = 1.4
• Sample size (\( n \)) = 85

Step 1: Calculate the standard error (SE)

As previously calculated: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{85}} \approx 0.151 \]

Step 2: Calculate the z-score for a sample mean of 2

Using the z-score formula: \[ z = \frac{\bar{x} - \mu}{SE} \] Where \( \bar{x} = 2 \), we have: \[ z = \frac{2 - 2.24}{0.151} \approx \frac{-0.24}{0.151} \approx -1.587 \]

Step 3: Find the corresponding probability

To find the probability of obtaining a z-score of less than -1.587, we can look this up in the standard normal distribution table or use a calculator:

From the previously calculated result or using a z-table:

• The cumulative probability for \( z = -1.587 \) is approximately \( P(Z < -1.587) \approx 0.0571 \).

Step 4: Analyze the result

Since \( P(Z < -1.587) \approx 0.0571 \), this probability is approximately 0.0571, which can also be expressed as 5.71%.

Conclusion

Since the probability is slightly above 5%, it indicates that getting a sample mean less than 2 is unusual but not extremely rare. In terms of statistical significance commonly used (like the 0.05 level), it can be considered unusual, but not exceptionally so.

Thus, the answer to whether it would be unusual for the sample mean to be less than 2 is:

\[ \text{Probability that sample mean < 2: } \approx 0.0571 \]

And, in terms of being unusual: \[ \text{It would be somewhat unusual for the sample mean to be less than 2 because the probability is } \boxed{0.0571}. \]

Answer 8

(e) Can you tell whether it is unusual for an individual household to have fewer than 2 TV sets? Explain. Assume the population is approximately normal.

Answer 9

To determine whether it is unusual for an individual household to have fewer than 2 TV sets, we can calculate the z-score for a household having 2 TV sets and find the corresponding probability. Since we assume the population is approximately normal, we can evaluate the position of the number of TV sets for an individual household.

Given:

• Population mean (\( \mu \)) = 2.24
• Population standard deviation (\( \sigma \)) = 1.4

Step 1: Calculate the z-score for 2 TV sets

Using the z-score formula: \[ z = \frac{X - \mu}{\sigma} \] Where \( X = 2 \) (the number of TV sets). Substituting the known values: \[ z = \frac{2 - 2.24}{1.4} \approx \frac{-0.24}{1.4} \approx -0.1714 \]

Step 2: Find the corresponding probability

Now we need to find the probability that the number of TV sets \( X \) is less than 2. We can find this using a standard normal distribution table or calculator to find the cumulative probability for \( z = -0.1714 \).

Using a z-table or calculator:

• The cumulative probability for \( z = -0.1714 \) is approximately \( P(Z < -0.1714) \approx 0.4312 \).

Step 3: Interpret the result

Since \( P(X < 2) \approx 0.4312 \), this means that approximately 43.12% of individual households would have fewer than 2 TV sets.

Conclusion

• Is it unusual for an individual household to have fewer than 2 TV sets?
  • A probability of 43.12% indicates that it is not unusual for an individual household to have fewer than 2 TV sets. In fact, it is quite common, given that almost half of the households fall into this category.

Thus, the conclusion would be: It is not unusual for an individual household to have fewer than 2 TV sets, as nearly 43.12% of households are expected to have this number or fewer.