SOLUTION: 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

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Question 863080: 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 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
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!
Found 2 solutions by ewatrrr, stanbon:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

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

2.294%2Fsqrt%2819%29 = .5263  Note: transposed digits(2.249 used)

TI, z = invNorm(.95) = 1.6449

1.6449 = %28X+-+93%29%2F%282.294%2Fsqrt%2819%29%29

X =.526*1.645 + 93 = 93.87

Answer by stanbon(75887) About Me  (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.
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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
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Cheers,
Stan H.