the amount of time spent in North American adults watching television per day is normal distributed with a mean of 6 hours and a standard deviation of 1.5 hours
What is your question?
When you add the values 3, 5, 8, 12, and 20 and then divide by the number of values, the result is 9.6. Which term best describes this value: average, mean, median, mode, or standard deviation?
To find probabilities of the amount of time spent watching television per day, we can use the properties of the normal distribution. Specifically, we can calculate z-scores and use a standard normal distribution table or a calculator to find the probabilities.
Step 1: Convert the given values to z-scores.
In this case, we want to find the probability of observing a certain amount of time spent watching TV. Let's say we want to find the probability of spending less than 4 hours watching TV.
First, we calculate the z-score using the formula:
z = (x - μ) / σ
where x is the observed value, μ is the mean, and σ is the standard deviation.
For our case, let's calculate the z-score for x = 4:
z = (4 - 6) / 1.5
Step 2: Look up the probability using a z-table or calculator.
Using the z-score we calculated in step 1, we can now find the probability associated with that z-score. The z-table provides the area under the standard normal distribution curve.
For example, if we want to find the probability of spending less than 4 hours watching TV, we look up the z-score of -2/1.5 in the z-table. The area associated with this z-score is the probability we seek.
Step 3: Interpret the result.
The probability you find will represent the proportion of North American adults who spend less than 4 hours watching TV per day.
To find probabilities and answer questions about a normal distribution, we need to use the z-score formula and the properties of the standard normal distribution.
The z-score formula is defined as:
z = (x - μ) / σ
Where:
- z is the z-score,
- x is the data point,
- μ is the mean of the distribution, and
- σ is the standard deviation of the distribution.
Now, let's use this information to answer some questions about the time spent watching television.
Question 1: What is the probability that a randomly selected North American adult watches TV for more than 5 hours?
Solution:
To answer this question, we need to calculate the z-score for 5 hours and then find the area to the right of that z-score.
z = (x - μ) / σ
z = (5 - 6) / 1.5
z = -0.67
Next, we need to find the area to the right of the z-score -0.67 in the standard normal distribution table (also known as the z-table). The z-table gives us the area under the curve to the left of a z-score. To find the area to the right, we subtract the value from 1.
Using the z-table, we find that the area to the left of -0.67 is 0.2514. Therefore, the area to the right is 1 - 0.2514 = 0.7486, or 74.86%.
So, the probability that a randomly selected North American adult watches TV for more than 5 hours is 74.86%.
Question 2: What is the probability that a randomly selected North American adult watches TV for less than 7 hours?
Solution:
To answer this question, we need to calculate the z-score for 7 hours and then find the area to the left of that z-score.
z = (x - μ) / σ
z = (7 - 6) / 1.5
z = 0.67
Next, we need to find the area to the left of the z-score 0.67 in the standard normal distribution table (z-table).
Using the z-table, we find that the area to the left of 0.67 is 0.7486.
So, the probability that a randomly selected North American adult watches TV for less than 7 hours is 74.86%.
Question 3: What is the probability that a randomly selected North American adult watches TV between 4 and 8 hours?
Solution:
To answer this question, we need to calculate the z-scores for 4 hours and 8 hours, and find the area between those two z-scores.
For 4 hours:
z = (x - μ) / σ
z = (4 - 6) / 1.5
z = -1.33
For 8 hours:
z = (x - μ) / σ
z = (8 - 6) / 1.5
z = 1.33
Next, we need to find the areas to the left of the z-scores -1.33 and 1.33 in the standard normal distribution table (z-table).
Using the z-table, we find that the area to the left of -1.33 is 0.0918 and the area to the left of 1.33 is 0.9082.
To find the area between -1.33 and 1.33, we subtract the area to the left of -1.33 from the area to the left of 1.33:
0.9082 - 0.0918 = 0.8164
So, the probability that a randomly selected North American adult watches TV between 4 and 8 hours is 81.64%.