In a factory producing glasses, the weights of glasses are known to have a mean of 160 grams. It is also known that the interquartile range of the weights of glasses is 28 grams. Assuming the weights of glasses to be normally distributed, find the standard deviation of the weights of glasses.

weight of glass = X

\(X \sim {\text{N}}(160,{\text{ }}{\sigma ^2})\)

\({\text{P}}(X < 160 + 14) = {\text{P}}(X < 174) = 0.75\)     (M1)(A1)

Note: \({\text{P}}(X < 160 - 14) = {\text{P}}(X < 146) = 0.25\) can also be used.

\({\text{P}}\left( {Z < \frac{{14}}{\sigma }} \right) = 0.75\)     (M1)

\(\frac{{14}}{\sigma } = 0.6745 \ldots \)     (M1)(A1)

\(\sigma = 20.8\)     A1

[6 marks]

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