To calculate the upper and lower quartiles of a normally distributed variable, we can use the mean and standard deviation of the distribution along with the properties of the normal distribution.

Given:

• Mean $\mu = 6 \, \text{days} = 144 \, \text{hours}$ (since 1 day = 24 hours)
• Standard deviation $\sigma = 7.5 \, \text{hours}$

The lower quartile (Q1) and upper quartile (Q3) correspond to the 25th and 75th percentiles of the distribution, respectively.

Using the standard normal distribution table:

• The z-score for the lower quartile (Q1) at the 25th percentile is approximately $-0.674$
• The z-score for the upper quartile (Q3) at the 75th percentile is approximately $+0.674$

Step 1: Calculate the lower quartile (Q1)

The formula to convert a z-score to a value $X$ in the original distribution is:

$X = \mu + (z \times \sigma)$

For Q1: $Q1 = 144 + (-0.674 \times 7.5) = 144 - 5.055 = 138.945 \, \text{hours}$ So, $Q1 \approx 138.95$ hours, which is approximately 5.79 days.

Step 2: Calculate the upper quartile (Q3)

For Q3: $Q3 = 144 + (0.674 \times 7.5) = 144 + 5.055 = 149.055 \, \text{hours}$ So, $Q3 \approx 149.06$ hours, which is approximately 6.21 days.

Summary:

• Lower quartile (Q1): 138.95 hours or 5.79 days
• Upper quartile (Q3): 149.06 hours or 6.21 days

Would you like a more detailed explanation or have any questions?

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Related Questions:

1. How would you calculate the median lifespan of these mosquitoes?
2. What is the probability that a mosquito lives less than 5 days?
3. How would the quartiles change if the standard deviation increased?
4. What is the lifespan range for the middle 50% of mosquitoes?
5. How can you interpret z-scores in real-world terms for this problem?

Tip: Remember that in a normal distribution, about 68% of the values fall within one standard deviation from the mean.