Normal distribution - statisticsThe thickness of some plastic shelving produced by a factory is normally distributed. As part of the production process the shelving is tested with two gauges. The first gauge is 7mm thick and 98.61% of the shelving passes through this gauge. The second gauge is 5.2mm thick and only 10.2% of the shelves pass through this gauge. Find the mean and standard deviation of the thickness of the shelvingI have created simultaneous equations and made one equal 2.20 and the other 2.30Any help would be much appreciated as i am really stuck on this oneThankyou very much

I am assuming that this is a standard normal distribution.

we have to find the `barx and s` , the mean and the standard deviation

we know for standard distribution,

`z = (x-barx)/s`

for x =7, percentile is 98.1 or 0.981

This gives the area between mean (z=0) and x=7 is (0.981-0.5) which is 0.481

From a Z-table (I have given the link below)

you get,

for x = 7, z = 2.05

for x = 5.2, percentile is 10.2 or 0.102

This gives the area between x =5.2 and mean(z=0) is (0.5 -0.102) which is 0.398

From a z table you get,

for x =5.2 z = -1.27

Now, using the above equation,

`2.05 = (7-barx)/s and -1.27 =(5.2 -barx)/s`

this gives you two equations,

`2.05s+barx = 7 and -1.27s+barx=5.2`

This gives,

`barx = 5.888 cm and s =0.542`

The mean is 5.888 cm(or 5.9 cm for accuracy) and standard deviation is 0.542.