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Temperatures in June in LA are distributed nearly normally with mean 77 degrees and standard deviation 5 degrees F. In what proportion of days in June in LA is the temperature over 85 degrees?

Respuesta :

jolis1796

Answer:

5.5%

Explanation:

Given

μ = 77 degrees

σ = 5 degrees

X = 85 degrees

First, we need to calculate the z-score

Z = (X-μ) / σ

Z = (85-77) / 5 = 1.60

Then, we need to refer to the normal probability table, locate 1.6 on the row and 0.00 on the column. And a Z-score of 1. 6 then corresponds to the value at the intersection, which is 0.9452.

Remember, the Z-score refer to the area under the curve below observation value, so to find the area above the curve below the observation value, we simply need to find the complement, which is 1 - 0.9452 = 0.0548, roughly the 5.5%.

raphealnwobi

5.48% of days in June in LA has temperature over 85 degrees.

z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\x=raw\ score,\mu=mean,\sigma=standard\ deviation[/tex]

Given that μ = 77, σ = 5

For x = 85:

z = (85 - 77)/5 = 1.6

From the normal distribution table, P(x > 85) = P(z > 1.6) = 1 - P(z < 1.6) = 1 - 0.9452 = 0.0548 = 5.48%

5.48% of days in June in LA has temperature over 85 degrees.

Find out more on z score at: https://brainly.com/question/25638875

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