Question2.
The mean June midday temperature in Desertville is $36^{\circ} \mathrm{C}$ and the standard deviation is $3^{\circ} \mathrm{C}$
Assuming this data is normally distributed, how many days in June would you expect the midday temperature to be between $39^{\circ} \mathrm{C}$ and $42^{\circ} \mathrm{C}$ ?
What temperature can you expect in 10\% warmest days in June?

Studdy Solution
Determine the z-score that corresponds to the top 10% of the distribution.
Find $z$ such that $P(Z < z) = 0.90$.
From the standard normal distribution table, $z \approx 1.28$.
Calculate the temperature corresponding to this z-score:
$$T = \mu + z \cdot \sigma = 36 + 1.28 \times 3 = 36 + 3.84 = 39.84^{\circ} \mathrm{C}$$
The expected number of days with temperatures between $39^{\circ} \mathrm{C}$ and $42^{\circ} \mathrm{C}$ is approximately 4 days. The temperature you can expect on the 10% warmest days in June is approximately $39.84^{\circ} \mathrm{C}$.