In a population of body-builders, it is found that 2011bs and 3.9, what is P (197<z<202)...

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Question

In a population of body-builders, it is found that 2011bs and 3.9, what is P (197<z<202) ? (Round to the nearest hundredth of

Answer 1

Answer #1

Solution :

Given that ,

mean = $\mu$ = 201

standard deviation = $\sigma$ = 3.9

P(197 $\leq$ x $\leq$ 202) = P((197 - 201 / 3.9) $\leq$ (x - $\mu$ ) / $\sigma$ $\leq$ (202 - 201 / 3.9) )

= P(-1.03 $\leq$ z $\leq$ 0.26)

= P(z $\leq$ 0.26) - P(z $\leq$ -1.03)

= 0.6026 - 0.1515 Using standard normal table,

Probability = 0.45

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Answer 2

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