Question

A set of average city temperatures in August are normally distributed with a mean of 21.25∘C21.25 ^\circ \text{C}21.25∘C21, point, 25, degrees, start text, C, end text and a standard deviation of 2∘C2 ^\circ \text{C}2∘C2, degrees, start text, C, end text.

What proportion of temperatures are between 23.71∘C23.71^\circ \text{C}23.71∘C23, point, 71, degrees, start text, C, end text and 26.17∘C26.17^\circ \text{C}26.17∘C26, point, 17, degrees, start text, C, end text?

You may round your answer to four decimal places.

Answer 1

Answer:$P(23.71<X<26.17)=P(\frac{23.71-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{26.17-\mu}{\sigma})=P(\frac{23.71-21.25}{2}<Z<\frac{26.17-21.25}{2})=P(1.23<z<2.46)$And we can find this probability with this difference:

$P(1.23<z<2.46)=P(z<2.46)-P(z<1.23)$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  $P(1.23<z<2.46)=P(z<2.46)-P(z<1.23)=0.9931-0.8907=0.1024$Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the temperaturs in August, and for this case we know the distribution for X is given by:

$X \sim N(21.25,2)$  

Where $\mu=21.25$ and $\sigma=2$

We are interested on this probability

$P(23.71<X<26.17)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(23.71<X<26.17)=P(\frac{23.71-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{26.17-\mu}{\sigma})=P(\frac{23.71-21.25}{2}<Z<\frac{26.17-21.25}{2})=P(1.23<z<2.46)$And we can find this probability with this difference:

$P(1.23<z<2.46)=P(z<2.46)-P(z<1.23)$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(1.23<z<2.46)=P(z<2.46)-P(z<1.23)=0.9931-0.8907=0.1024$

Answer 2

Using the normal distribution, it is found that 0.1024 = 10.24% of temperatures are between 23.71ºC and 26.17ºC.

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Normal Probability Distribution

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. Looking at the z-score table and find the p-value associated with this z-score, which is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

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• Mean of 21.25ºC means that $\mu = 21.25$
• Standard deviation of 2ºC means that $\sigma = 2$
• The proportion of temperatures between 23.71ºC and 26.17ºC is the p-value of Z when X = 26.17 subtracted by the p-value of Z when X = 23.71. Thus:

X = 26.17

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{26.17 - 21.25}{2}$

$Z = 2.46$

$Z = 2.46$ has a p-value of 0.9931.

X = 23.71

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{23.71 - 21.25}{2}$

$Z = 1.23$

$Z = 1.23$ has a p-value of 0.8907.

0.9931 - 0.8907 = 0.1024

0.1024 = 10.24% of temperatures are between 23.71ºC and 26.17ºC.

A similar problem is given at https://brainly.com/question/15017791