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Dafna1 [17]
3 years ago
15

The distribution of the amount of money spent by first-time gamblers at a major casino in las vegas is approximately normal in s

hape 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?
Mathematics
1 answer:
Lostsunrise [7]3 years ago
5 0

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.

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