VIPSolutionsGive step-by-step solution with explanation and final answer: The average time a subscriber spends reading the local newspaper is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. For the 10% who spend the most time reading the paper, how much time do they spend? a. 11.72 b. 28.52 C. 86.28 d. 69.48
Question:
Give step-by-step solution with explanation and final answer: The average time a subscriber spends reading the local newspaper is 49 minutes. Assume the standard deviation is 16 minutes and that the times are normally distributed. For the 10% who spend the most time reading the paper, how much time do they spend? a. 11.72 b. 28.52 C. 86.28 d. 69.48
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Asked by: nayan
Created at: 2025-09-17 17:23:30
Asked by: nayan
Created at: 2025-09-17 17:23:30
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## Given Data
- Mean (\(\mu\)): 49 minutes
- Standard deviation (\(\sigma\)): 16 minutes
- Distribution: Normal
- Top 10%: Find the minimum time (\(x\)) that puts a reader in the top 10% of time spent.
---
## Step 1: Find the Z-score for the Top 10%
The top 10% corresponds to the **90th percentile** (since 90% are below, 10% are above).
From the standard normal table, the Z-score for the 90th percentile is approximately:
\[
z = 1.28
\]
---
## Step 2: Convert Z-score to Raw Score
The formula for converting a Z-score to an actual value (\(x\)) is:
\[
x = \mu + z\sigma
\]
---
## Step 3: Plug in the Values
\[
x = 49 + (1.28)(16)
\]
\[
x = 49 + 20.48
\]
\[
x = 69.48
\]
---
## Step 4: Final Answer
**For the 10% who spend the most time reading the paper, they spend at least:**
\[
\boxed{69.48 \text{ minutes}}
\]
**Answer:** (d) 69.48
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