# Hey guys I need some help! 1)Water pressure problems occur in the system after 99% capacity is reached. On what percentage of the days were there problems? mean=87 standard deviation=11.4704...

**Hey guys I need some help!**

**1)Water pressure problems occur in the system after 99% capacity is reached. On what percentage of the days were there problems? **

**mean=87**

**standard deviation=11.4704**

**lower bound=99**

**upper bound=large number, **

**Using the normal distribution feature on the calculator, **

**normalcdf(99,1E99,87,11.4704)= 0.1477 or, **

**Using z>1.04617 normalcdf(1.04617, 5)**

**The area under the curve would be 0.1477.**

**Therefore there were problems on 14.77% of the days.**

***My question is what is ****1E99 and why is expressed**** like this?**

***next question! from ****where they got****that result****Using z>1.04617 normalcdf(1.04617, 5)**

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(1) 1E99 is engineering notation for `1"x"10^99` , or effectively infinity. Your calculator is finding the area under the standard normal curve between two values. It uses a numerical approximation technique, and is unable to handle infinity as an input. However, it can handle an absurdly large number like `10^99` .

(2) In a TI-83/84 normalcdf(99,1E99,87,11.4704) gives the area under the standard normal curve from 99 to infinity* if the mean is 87 and the standard deviation is 11.4704.

Alternatively, you could convert all the numbers to standard z-scores. If `x=99,mu=87,sigma=11.4704` then `z=(99-87)/(11.4704)~~1.0462`

Then P(x>99)=P(z>1.0462) and the latter can be found by using normalcdf(1.0462,1E99) as this assumes a mean of 0 and a standard deviation of 1.

Hope this helps.