Time Spent Online Americans spend an average of 5 hours per day online. If the standard deviation is 29 minutes, find the range in which at least 93.75% of the data will lie. Use Chebyshev's theorem. Round your k to the nearest whole number. At least 93.75% of the data will lie between and minutes.
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
![### Time Spent Online
Americans spend an average of 5 hours per day online. If the standard deviation is 29 minutes, find the range in which at least 93.75% of the data will lie. Use Chebyshev's theorem. Round your \( k \) to the nearest whole number.
**Solution:**
To determine the range, apply Chebyshev's theorem, which states that:
\[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \]
Given:
- Mean (\( \mu \)) = 5 hours (300 minutes)
- Standard deviation (\( \sigma \)) = 29 minutes
- At least 93.75% of the data (\( P \)) = 0.9375
First, find \( k \):
\[ 0.9375 = 1 - \frac{1}{k^2} \]
\[ \frac{1}{k^2} = 1 - 0.9375 \]
\[ \frac{1}{k^2} = 0.0625 \]
\[ k^2 = \frac{1}{0.0625} \]
\[ k = \sqrt{\frac{1}{0.0625}} \]
\[ k = \sqrt{16} \]
\[ k = 4 \]
Now, determine the range:
\[ \mu - k\sigma \] and \[ \mu + k\sigma \]
Calculate both bounds:
\[ 300 - 4(29) \] and \[ 300 + 4(29) \]
\[ 300 - 116 \] and \[ 300 + 116 \]
\[ 184 \] and \[ 416 \]
Therefore, at least 93.75% of the data will lie between 184 and 416 minutes.
* **Input Fields:**
- Between: \( \boxed{184} \) and \( \boxed{416} \) minutes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb63fc9e3-7584-4c90-be8d-bc130c8a2c55%2F46154d7e-a464-4dfd-b8a6-5a23ed61214f%2Fyx1cmdm_processed.jpeg&w=3840&q=75)
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