Question

The number of cigarettes smoked per day by adults forms a normal distribution. The average number of cigarettes smoked per day is 20 with a standard deviation of 7. What number of cigarettes would represent the 35th percentile?

Answer 1

Answer:

17.3 cigarettes represent the 35th percentile

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 20, \sigma = 7$

What number of cigarettes would represent the 35th percentile?

This is the value of X when Z has a pvalue of 0.35. So it is X when Z = -0.385.

So

$Z = \frac{X - \mu}{\sigma}$

$-0.385 = \frac{X - 20}{7}$

$X - 20 = -0.385*7$

$X = 17.3$

17.3 cigarettes represent the 35th percentile