im not sure how to do this question

Question

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.933 g and a standard deviation of 0.308 g. Find the probability of randomly selecting a cigarette with 0.502 g of nicotine or less. P(X < 0.502 g)

 

Answer 1

Hi Meili,

To use the normal distribution, you must convert the raw score of .502 to a z-score by using the formula z = (x - μ)/σ So

                                            z = (.502 - .933)/.308 = -1.399.....

If you are using a table that is in your textbook, you will have to round this number to -1.40.  Therefore P(x<0.502)=P(z<-1.40)=.0808