Given a normal distribution of values for which the mean is 70 and the standard deviation is 4.5. Find the probability that a value is less than 62

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Given mean `mu=70`

Standard deviation `sigma=4.5`

`P(xlt62)=P((x-mu)/sigmalt(62-70)/4.5)`

` =p(zlt-1.777)`

` =0.0377`

From a standard normal table we find that the probability of a z-value less than -1.777 is 0.0377.

**Therefore, the probability that a value is less than 62 is approximately 3.77 percent.**

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