# Normal distribution - statistics The 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 shelving I have created simultaneous equations and made one equal 2.20 and the other 2.30 Any help would be much appreciated as i am really stuck on this one Thankyou 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...

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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.

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