Answer:
The probability is [tex]P(X \le 0.305 ) = 0.01795[/tex]
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu = 0.966 \ grams[/tex]
The standard deviation is [tex]\sigma = 0.315 \ grams[/tex]
Given that the amounts of nicotine in a certain brand of cigarette are normally distributed
Then the probability of randomly selecting a cigarette with 0.305 grams of nicotine or less is mathematically represented as
[tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P(\frac{X - \mu }{\sigma } > \frac{0.305 - \mu }{\sigma } )[/tex]
Generally
[tex]\frac{X - \mu }{\sigma } = Z (The \ standardized \ value \ of X )[/tex]
So
[tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P(Z > \frac{0.305 - 0.966 }{0.315} )[/tex]
[tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - P(Z >-2.0984 )[/tex]
From the z-table(reference calculator dot net ) value of [tex]P(Z >-2.0984 ) =0.98205[/tex]
So
[tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 1 - 0.98205[/tex]
=> [tex]P(X \le 0.305 ) = 1 - P(X > 0.305) = 0.01795[/tex]
=> [tex]P(X \le 0.305 ) = 0.01795[/tex]
