Chi square test

[schdero]

A producer of a certain good claims that the durability of his goods is normally distributed. In a sample of n=90 , the average durability x=3.3 with a standard deviation of s=0.6. The sample shows the following distribution
Durability in years Number of individual goods
0-2.5 6
2.5-3 8
3-3.5 35
3.5-4 25
4-4.5 10
>4.5 6

Test with a Chi square test whether or not the claim of the producer is valid.
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My question is about the calculation of the expected number of individual goods in a certain class (eg 0-2.5). To do that one needs to apply the z-value table after standardizing the distribution. For 2.5 e.g., the z value is (according to the official solution) (2.5-3.3)/0.6. However, what i dont get is why it is not (90)^0.5*(2.5-3.3)/0.6.

Does anybody know why you do not include the size of the sample but only the standard deviation of it?

Thank you

Schdero

[David1]

The expectation includes the the size of the sample
E=P(a<x<b)*n
but the probability itself doesn't, since you are asked to calculate the probability that x will be in between a and b, not the probability that \overline{x} will be in between a and b.