Question

Math

Posted 5 months ago

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A set of average city temperatures in July are normally distributed with a mean of $23.5^{\circ} \mathrm{C}$ and a standard deviation of $2^{\circ} \mathrm{C}$. The average temperature of Rabat is $23^{\circ} \mathrm{C}$.
What proportion of average city temperatures are higher than that of Rabat?
You may round your answer to four decimal places.
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

Calculate the z-score for the given x value

step 2

The z-score is calculated using the formula $z = \frac{x - \mu}{\sigma}$ where $x$ is the value in question, $\mu$ is the mean, and $\sigma$ is the standard deviation

step 3

For $x = 23$, $\mu = 23.5$, and $\sigma = 2$, the z-score is $z = \frac{23 - 23.5}{2} = -0.25$

step 4

Use a standard normal distribution table or calculator to find the probability that $Z > -0.25$

step 5

The probability that $Z > -0.25$ corresponds to the proportion of city temperatures higher than that of Rabat

step 6

This probability is $0.5 + P(0 < Z < 0.25)$, since the total area under the normal curve is 1 and the curve is symmetric about the mean

step 7

Using standard normal distribution tables or a calculator, $P(0 < Z < 0.25)$ is approximately 0.0987

step 8

Therefore, the proportion of city temperatures higher than that of Rabat is $0.5 + 0.0987 = 0.5987$

step 9

Round the answer to four decimal places as requested

Answer

0.5987

Key Concept

Calculating the proportion of values above a given point in a normal distribution using the z-score

Explanation

The z-score represents the number of standard deviations a value is from the mean. The area to the right of the z-score in a standard normal distribution gives the proportion of values above that value.

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