# 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 gotthat resultUsing z>1.04617 normalcdf(1.04617, 5)

Posted on

(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.

We’ve answered 323,666 questions. We can answer yours, too.