Answer :
Answer: 0.3292
Step-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
Answer:0.3254
Step-by-step explanation:
Using normal distribution
z = (x - mean) / standard deviation
Where mean =72.5
Standard deviation =2.3
z = standard normal variable
x = values of temperature of the lake
We are looking for probability that the temperature of the lake is between 71 and 73 degrees
=P( 71 lesser than/equal to x lesser than/equal to 73
For x= 71,
z = (71-72.5)/2.3= -0.6522
For x= 73,
z = (73-72.5)/2.3= 0.2174
Looking at the normal distribution table
For area covered by (considering z=-0.6522)
(z = -0.6522 to z= 0) =0.2578
For area covered by (considering z =0.2174)
(z = 0 to z= 0.2174) = 0.5832
P( 71 lesser than/equal to x lesser than/equal to 73)
= 0.5832-0.2578= 0.3254
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