Q:

During the month of August, the average temperature of a lake next to a local college is 72 . 5 degrees with a standard deviation of 2 . 3 degrees. For a randomly selected day, find the probability that that the temperature of the lake is between 71 and 73 degrees. (Assume normal distribution)

Accepted Solution

A:
Answer: 0.3292Step-by-step explanation:Let x be the random variable that represents the temperature of a lake.Given :  [tex]\mu=72.5[/tex] and [tex]\sigma=2.3[/tex].Using formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]We assume that the  temperature of a lake follows a normal distribution.Z-score corresponds to x= 71[tex]z=\dfrac{71-72.5}{2.3}=-0.65217391304\approx-0.65[/tex]Z-score corresponds to x= 73[tex]z=\dfrac{73-72.5}{2.3}=0.217391304348\approx0.22[/tex]The probability that that the temperature of the lake is between 71 and 73 degrees :[tex]P(71<x<73)=P(-0.65<z<0.22)\\\\=P(z<0.22)-P(z<-0.65)\\\\=P(z<0.22)-(1-P(z<0.65))\\\\=0.5870644-(1-0.7421538)[/tex] [using z-table for right tailed test][tex]=0.3292182\approx0.3292[/tex]Hence, the probability that that the temperature of the lake is between 71 and 73 degrees =0.3292