Hi

Re TY, 94.6 is not going to work...

 = .5263  Note: transposed digits(2.249 used)

TI, z = invNorm(.95) = 1.6449

1.6449 = 

X =.526*1.645 + 93 = 93.87

 

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
 There is a 5% chance that the mean reading speed of a random samlple of 19 second grade studetns will exceed what value? u(x) =93 and SD = 2.294 

Use technology to find the sample mean x-bar for which there is an area of 0.05 to the right of the normal curve with u(x) of 93 and SD 2.294 

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I would like to know how to find the answer using a TI83 calculator. The answer is 94.6 and I have no idea how that is!

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Note:: Hard to tell if the given SD is for the sample means or for the

population of reading speeds.

If if is the SD for the means use 2.294/sqrt(19)

If it is the SD for the population use  2.294

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I assumed it was the SD for the population.

================================================

Find the z-value with a left tail of 0.95::

invNorm(0.95) = 1.645

Find the x-bar value using x-bar = z*s+u

x-bar = 1.645*(2.294/sqrt(19))+93 = 93.86 

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If you assume the SD = 2.294 if the SD of the sample means

you get x-bar = 1.645*2.294+93 = 96.8

===================

Cheers,

Stan H.



 

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