Mathematics: The mean June midday temperature in Desertville is 36°C and the standard deviation is 3°C.Assuming this data is normally distributed, how many days in June would you expect the midday temperature to be between 39°C and 42°C?

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The mean June midday temperature in Desertville

mega29

09.02.2022

The value is $E(X) = 4 \ days$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 36^oC$

The standard deviation $\sigma = 3^oC$

Generally the probability that in June , the midday temperature is between

39°C and 42°C is mathematically represented as

$P(39 < X < 42) = P(\frac{39 - 36}{3} < \frac{X - \mu }{\sigma} < (\frac{42 - 36}{3} )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

P(39 < X < 42) = P(1 < Z

=> P(39 < X < 42) = P(Z < 2) - P( Z

From the z table the area under the normal curve to the left corresponding to 1 and 2 is

P(Z < 2) = 0.97725

and

P(Z < 1) = 0.84134

P(39 < X < 42) = 0.97725 - 0.84134

=> P(39 < X < 42) = 0.13591

Generally number of days in June would you expect the midday temperature to be between 39°C and 42°C

E(X) = n * P(39 < X 42 )

Here n is the number of days in June which is n = 30

E(X) = 30 * 0.13591

=> $E(X) = 4 \ days$

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