mean = 36
sd = 3
P( 39 < x < 42) = P(( 39 - 36) / 3 < Z < ( 42 - 36) / 3)
P(3/3 <z<6/3)
P( 1 < Z < 2) = P( x < 2) - P( x < 1) = 0.9772 - 0.8413 = 0.1359
(need to use the normal probability table to get the decimal numbers)
June has 30 days
0.1359*30 = 4.0777 days, round to 4 days
The number of days in June when the midday temperature was between 39ºC and 42ºC is 4.
A z-table also known as the standard normal distribution table, helps us to know the percentage of values that are below (or to the left of the Distribution) a z-score in the standard normal distribution.
We know that the mean midday temperature recorded in June in a city in southern California is 36ºC(μ), and the standard deviation is 3ºC(σ). Also, it is given that the data is normally distributed, therefore,
P(X₁ ≤ X ≤ X₂)
=P(1 ≤ Z ≤ 2)
= P(Z ≤ 2) - P(Z ≤ 1)
= 0.9772 - 0.8413
= 0.1359
= 13.9%
Now, as we know the percentage of days that will have the temperature between 39ºC and 42ºC, and the month of June has 30 days. therefore, the number of days that will have the midday temperature was between 39ºC and 42ºC are,
Number of days = Total number of days is June × Percentage of days
= 30 × 13.59%
= 4.077 ≈ 4
Hence, the number of days in June when the midday temperature was between 39ºC and 42ºC is 4.
Learn more about Z-table:
https://brainly.com/question/6096474
