Using the normal distribution, it is found that 0.2119 = 21.19% of rainy days have rainfall with pH below 5.0.
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 5.43, \sigma = 0.54[/tex]
The proportion of rainy days have rainfall with pH below 5.0 is the p-value of Z when X = 5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 5.43}{0.54}[/tex]
Z = -0.8
Z = -0.8 has a p-value of 0.2119.
0.2119 = 21.19% of rainy days have rainfall with pH below 5.0.
More can be learned about the normal distribution at https://brainly.com/question/25800303
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