Assume that among the members of a men's gym, the distribution of body weights has a mean of 188 pounds and a standard deviation of 7. If 267 men belong to the gym, how many of them do you expect to be over 200 pounds?

Question

amistre64 · Answer

we have an average; and a deviation; and a goal

amistre64 · Answer

$$z=\frac{x-\bar x}{sd}$$

$$z=\frac{200-188}{7}$$

amistre64 · Answer

we then check the ztables to see how that zscore relates to a probability

amistre64 · Answer

|dw:1334776338026:dw|

anonymous · Answer

I come up with 1.71  Is that even close?

amistre64 · Answer

for the zscore?  maybe

1.714  looks fine; now we need to get a ztable

amistre64 · Answer

http://www.wolframalpha.com/input/?i=zscore+%28200-188%29%2F7 this is simpler tho; but it should be a more accurate value from the wolf than you would get from the table

anonymous · Answer

I'm clicking on the link now

amistre64 · Answer

the value from that that we want is the .04324

amistre64 · Answer

thats the area to the right of our z, and the pic i drew represents that area to the right of the z

amistre64 · Answer

so take the number of men: 267

and multiply it by .04324 and round up to the next whole number

anonymous · Answer

Thanks again. You really helped me understand this. And thank goodness for the TI-83!!