Mathematics: The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.966 grams and a standard deviation of 0.315 grams. Find the probability of randomly selecting a cigarette with 0.305 grams of nicotine or less.

Here s the answer JUST correct me if I m wrong Step-by-step explanation:Btw brainliest me plss...

The amounts of nicotine in a certain brand of

BrookLPrescott287

31.01.2022

The probability is $P(X \le 0.305 ) = 0.01795$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.966 \ grams$

The standard deviation is $\sigma = 0.315 \ grams$

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

$P(X \le 0.305 ) = 1 - P(X 0.305) = 1 - P(\frac{X - \mu }{\sigma } \frac{0.305 - \mu }{\sigma } )$

Generally

$\frac{X - \mu }{\sigma } = Z (The \ standardized \ value \ of X )$

$P(X \le 0.305 ) = 1 - P(X 0.305) = 1 - P(Z \frac{0.305 - 0.966 }{0.315} )$

$P(X \le 0.305 ) = 1 - P(X 0.305) = 1 - P(Z -2.0984 )$

From the z-table(reference calculator dot net ) value of P(Z -2.0984 ) =0.98205

$P(X \le 0.305 ) = 1 - P(X 0.305) = 1 - 0.98205$

=> $P(X \le 0.305 ) = 1 - P(X 0.305) = 0.01795$

=> $P(X \le 0.305 ) = 0.01795$

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