Question

The distribution of the amount of money spent by first-time gamblers at a major casino in las vegas is approximately normal in shape with a mean of 600 a n d a s t a n d a r d d e v i a t i o n o f 120. according to the standard deviation rule, almost 84% of gamblers spent more than what amount of money at this casino?

Answer 1

Solution: We are given:

$\mu=600, \sigma =120$

We need to find the z value corresponding to probability 0.84, in order to find the how much money almost 84% of gamblers spent at casino.

Using the standard normal table, we have:

$z(0.85) = 0.9945$

Now we will use the z score formula to find the required amount:

$z=\frac{x-\mu}{\sigma}$

$0.9945=\frac{x-600}{120}$

$0.9945 \times 120 = x - 600$

$119.34 = x - 600$

$x = 600 + 119.34$        

$x = 719.34$

$x = 720$ approximately

Therefore, almost 84% of gamblers spent more than $720 amount of money at this casino.