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

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

=

Answer 2

Answer: 0.3292

Step-by-step explanation:

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

Given :  $\mu=72.5$ and $\sigma=2.3$.

Using formula : $z=\dfrac{x-\mu}{\sigma}$

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

Z-score corresponds to x= 71

$z=\dfrac{71-72.5}{2.3}=-0.65217391304\approx-0.65$

Z-score corresponds to x= 73

$z=\dfrac{73-72.5}{2.3}=0.217391304348\approx0.22$

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

$P(71$ [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