Question

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)

Answer 1

Answer:0.3254

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

=

Answer 2

Answer: 0.3292

Explanation:

Let x be the random variable that represents the temperature of a lake.

Given :
`\mu=72.5` and
`\sigma=2.3`.

Using formula :
`z=(x-\mu)/(\sigma)`

We assume that the temperature of a lake follows a normal distribution.

Z-score corresponds to x= 71

`z=(71-72.5)/(2.3)=-0.65217391304\approx-0.65`

Z-score corresponds to x= 73

`z=(73-72.5)/(2.3)=0.217391304348\approx0.22`

The probability that that the temperature of the lake is between 71 and 73 degrees :

`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)` [using z-table for right tailed test]

`=0.3292182\approx0.3292`

Hence, the probability that that the temperature of the lake is between 71 and 73 degrees =0.3292