Given that the daily screen time of students follows an approximate normal distribution, we need to determine the mean $\mu$ of the distribution using the information provided for two students, Justin and Nasir, without using a z-score chart.

Step-by-step breakdown:

1. Justin’s Screen Time (Top 5%):

   • Justin spends 245 minutes on his phone each day, and this places him in the top 5% of the distribution.
   • In a normal distribution, the top 5% corresponds to a z-score of approximately $z = 1.645$ because 95% of the data falls below this value.
   • This means that Justin's screen time corresponds to a z-score of $z_J = 1.645$.

2. Nasir’s Screen Time (Bottom 15%):

   • Nasir spends 98 minutes on his phone each day, placing him in the bottom 15% of the distribution.
   • For the bottom 15%, the corresponding z-score is approximately $z = -1.036$ because the z-score that leaves 15% of the data to the left is $-1.036$.
   • This means Nasir's screen time corresponds to a z-score of $z_N = -1.036$.

Using the z-score formula:

The z-score formula is:

$z = \frac{x - \mu}{\sigma}$

Where:

• $x$ is the observed value (screen time),
• $\mu$ is the mean,
• $\sigma$ is the standard deviation of the distribution.

Using this formula, we can set up two equations based on the given information:

1. For Justin: $1.645 = \frac{245 - \mu}{\sigma}$ (Equation 1)

2. For Nasir: $-1.036 = \frac{98 - \mu}{\sigma}$ (Equation 2)

Solving the system of equations:

Now, we solve these two equations simultaneously.

Step 1: Solve each equation for $\sigma$:

From Equation 1: $\sigma = \frac{245 - \mu}{1.645}$

From Equation 2: $\sigma = \frac{98 - \mu}{-1.036}$

Step 2: Set the two expressions for $\sigma$ equal to each other:

$\frac{245 - \mu}{1.645} = \frac{98 - \mu}{-1.036}$

Step 3: Cross-multiply and solve for $\mu$:

$(245 - \mu)(-1.036) = (98 - \mu)(1.645)$

Expanding both sides: $-1.036 \times 245 + 1.036 \mu = 1.645 \times 98 - 1.645 \mu$ $-253.82 + 1.036 \mu = 161.21 - 1.645 \mu$

Step 4: Combine like terms:

$1.036 \mu + 1.645 \mu = 161.21 + 253.82$ $2.681 \mu = 415.03$

Step 5: Solve for $\mu$:

$\mu = \frac{415.03}{2.681} \approx 154.81$

Thus, the mean screen time $\mu$ for students at the University of Michigan is approximately 154.81 minutes per day.

Would you like any more details or explanations on this solution?

Related questions:

1. How can you calculate the standard deviation of the screen time distribution using the provided information?
2. If another student reports spending 200 minutes a day on their phone, what percentile would they be in?
3. How does the z-score value relate to percentiles in a normal distribution?
4. What are the characteristics of a normal distribution that make problems like this solvable?
5. How would the problem change if the distribution of screen times was not normal?

Tip:

Understanding the relationship between z-scores and percentiles in a normal distribution allows for easier interpretation of data points within any distribution.