# Normal Probability distribution?

• October 20th 2010, 08:57 AM
sankeel
Normal Probability distribution?
It is estimated that among customer type A the net profit is normally distributed with mean 450 and stdv 50.

If we do a box plot of net profit for a large sample of performance seekers, what's the proportion of performance seekers will be outliers according to the box plot?

upper inner fence = third quartile + 1.5(interquartile range)
lower inner fence = first quartile + 1.5(interquartile range)

The third quartile is 0.67 x stdv. But I don't know why?
• October 20th 2010, 08:37 PM
pickslides
Quote:

Originally Posted by sankeel

The third quartile is 0.67 x stdv. But I don't know why?

For 3rd quartile

$P\left( Z> a\right) = 0.75 \implies a = 0.6745$

Now $0.6745 = \frac{x-450}{50}$

$0.6745\times 50 = x-450$

$0.6745\times 50+450 = x$

$x = 484$