Question

Math

Posted 3 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 3 months ago

Solution by Steps

step 1

Calculate the z-score for Rabat's average temperature

step 2

The z-score is calculated using the formula $z = \frac{X - \mu}{\sigma}$, where $X$ is Rabat's temperature, $\mu$ is the mean temperature, and $\sigma$ is the standard deviation

step 3

Substituting the given values, we get $z = \frac{23 - 23.5}{2} = -0.25$

step 4

Look up the z-score in the standard normal distribution table or use a calculator to find the proportion of the area to the right of this z-score

step 5

The area to the right of $z = -0.25$ corresponds to the proportion of average city temperatures higher than that of Rabat

step 6

Using the standard normal distribution table or a calculator, we find that the area to the right of $z = -0.25$ is approximately 0.5987

Answer

The proportion of average city temperatures higher than that of Rabat is approximately 0.5987.

Key Concept

Z-score and Normal Distribution

Explanation

The z-score represents the number of standard deviations a data point is from the mean. In a normal distribution, the area to the right of a negative z-score represents the proportion of data points that are higher than the given value

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