Answer: C. 0.80920
Step-by-step explanation:
Given : The average low temperature in the park for May follows a normal distribution with a mean of [tex]\mu=32^{\circ}F[/tex] and a standard deviation of [tex]s=8^{\circ}F[/tex].
Let x represents the temperature in the park for May .
Using formula : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x= 25
[tex]z=\dfrac{25-32}{8}=-0.875[/tex]
Then by using the standard z-table for right tail test, [tex]P(x>25)=P(z>-0.875)\\\\=1-P(z>0.875)\\\\=1-0.1908=0.80920[/tex]
Hence, the probability that this bag will be warm enough on a randomly selected May night at the park= 0.80920