Little's Law

We can use Little's law $N=rt$ to solve this problem

It was given that ,

• The store owner estimates that there are 45 shoppers/buyers at the store at any time

• For the new store, the owner estimates that during business hours, an average of 90 shoppers per hour

• Each of 90 shoppers stays an average of 12 minutes

Now , using Little's law

$N=rt$

= 1.5 $×$ 12 ;[Since 90 shoppers per hour is equivalent to 1.5 shoppers per minute]

= 18

18 shoppers In the new store at any time

So, $\frac{45 - 18}{45}$ $×$ 100 = 60 %

According to the original information given, the estimated average number of shoppers in the original store at any time is 45. In the question, it states that, in the new store, the manager estimates that an average of 90 shoppers per hour enter the store, which is equivalent to 1.5 shoppers per minute. The manager also estimates that each shopper stays in the store for an average of 12 minutes. Thus, by Little’s law, there are, on average, $N=rt=(1.5)(12)=18$ shoppers in the new store at any time. This is $\dfrac{45-18}{45}*100=60$ percent less than the average number of shoppers in the original store at any time.

The final answer is $\large\color{#69047E}\boxed{60}$ .