Answer:
0.2855
Step-by-step explanation:
Mean of the daily water level of the lake=M=84.07
The probability that the daily water level in the lake is at least 100 feet is 0.064. We can write it as P(X≥100)=0.064 In a normal distribution corresponding z-score of X to this probability is z= 1.52204
From here, standart deviation of the distribution s can be found by using the formula:
z= [tex]\frac{X-M}{s}[/tex] i.e
s= [tex]\frac{X-M}{z}[/tex] = 10.4662
For P(X≥90), we can calculate z-score of X:
[tex]\frac{90-84.07}{10.4662}[/tex] =0.5666
Using this z value P(X≥90) = 0.2855