Respuesta :
Answer:
11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June
The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 77, \sigma = 5[/tex]
What is the probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June?
This probability is 1 subtracted by the pvalue of Z when X = 83. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{83 - 77}{5}[/tex]
[tex]Z = 1.2[/tex]
[tex]Z = 1.2[/tex] has a pvalue of 0.8849.
1 - 0.8849 = 0.1151
11.51% probability of observing an 83 degree F temperature or higher LA during a randomly chosen day in June
How cold are the coldest 10% of the days during June in LA?
High temperatures of X or lower, in which X is found when Z has a pvalue of 0.1, so whn Z = -1.28
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{X - 77}{5}[/tex]
[tex]X - 77 = -1.28*5[/tex]
[tex]X = 70.6[/tex]
The coldest 10% of the days during June in LA have high temperatures of 70.6F or lower.
A) The probability of observing a temperature ≥ 83°F in LA during a randomly chosen day in June is;
p(observing a temperature ≥ 83°F) = 11.507%
B) The coldest 10% of the days during June in LA have temperatures;
Less than or equal to 70.592 °F
This question involves z-distribution which is given by the formula;
z = (x' - μ)/σ
We are given;
Average daily temperature; μ = 77 °F
Standard deviation; σ = 5 °F
Since the temperatures follow a normal distribution, then if we want to find the probability of observing a temperature ≥ 83°F, then;
x' = 83 °F
Thus;
z = (83 - 77)/5
z = 6/5
z = 1.2
Thus;
from online z-score calculator, p-value = 0.11507
Thus, p(observing a temperature ≥ 83°F) = 11.507%
B) We want to find out how cold the coldest 10% of the days during June in LA;
Thus, it means that p = 10% = 0.1
z-score at p = 0.1 from z-score tables is;
z = -1.28155
Thus;
-1.28155 = (x' - 77)/5
-1.28155*5 = x' - 77
-6.40775 = x' - 77
x' = 77 - 6.40775
x' ≈ 70.592 °F
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