Question

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?

Answer 1

Answer:

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

       $E(X) = 30 * 0.13591$

=>    $E(X) = 4 \ days$