Question 1179878: The daily temperature in February is normally distributed with a mean of 25 degrees and a standard deviation of 12 degrees. What percentage of the time will this temperature be:
A. above freezing? (32 degrees or more)
B. below zero?
c. between 35 degrees and 45 degrees?
Explain your answers.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve each part of this problem step-by-step:
Given:
Mean (μ) = 25 degrees
Standard deviation (σ) = 12 degrees
Distribution: Normal
A. Above Freezing (32 degrees or more):
Calculate the z-score:
z = (x - μ) / σ
z = (32 - 25) / 12
z = 7 / 12 ≈ 0.58
Find the probability:
We want P(X ≥ 32), which is the same as P(Z ≥ 0.58).
Using a z-table or calculator, find P(Z < 0.58) ≈ 0.7190.
P(Z ≥ 0.58) = 1 - P(Z < 0.58) = 1 - 0.7190 = 0.2810
Convert to percentage:
0.2810 * 100% = 28.10%
Explanation: There's about a 28.10% chance that the daily temperature will be above freezing (32 degrees or more).
B. Below Zero:
Calculate the z-score:
z = (x - μ) / σ
z = (0 - 25) / 12
z = -25 / 12 ≈ -2.08
Find the probability:
We want P(X < 0), which is the same as P(Z < -2.08).
Using a z-table or calculator, find P(Z < -2.08) ≈ 0.0188.
Convert to percentage:
0.0188 * 100% = 1.88%
Explanation: There's about a 1.88% chance that the daily temperature will be below zero.
C. Between 35 degrees and 45 degrees:
Calculate the z-scores:
For 35 degrees:
z1 = (35 - 25) / 12 = 10 / 12 ≈ 0.83
For 45 degrees:
z2 = (45 - 25) / 12 = 20 / 12 ≈ 1.67
Find the probability:
We want P(35 ≤ X ≤ 45), which is the same as P(0.83 ≤ Z ≤ 1.67).
P(0.83 ≤ Z ≤ 1.67) = P(Z ≤ 1.67) - P(Z ≤ 0.83)
Using a z-table or calculator:
P(Z ≤ 1.67) ≈ 0.9525
P(Z ≤ 0.83) ≈ 0.7967
P(0.83 ≤ Z ≤ 1.67) = 0.9525 - 0.7967 = 0.1558
Convert to percentage:
0.1558 * 100% = 15.58%
Explanation: There's about a 15.58% chance that the daily temperature will be between 35 and 45 degrees.
Answers:
A. 28.10%
B. 1.88%
C. 15.58%
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