The scores on a standardized test in a suburban high school have a mean of 80 with a standard deviation of 12. What is the probability that a student will have a score less than 60.
Also, the average time to complete the test in the above question) is 40 minutes with a standard deviation of 5.2 minutes. find the probability that a student will requrie more than 50 minutes to finish the test.
Originally Posted by wild_flowr69
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I'm including an explanation of z-sxores with different examples. Please see how far you can proceed with your questions and then write back with the steps you've taken if you're still stuck.
Z-scores are a way of counting standard deviations from the mean. So 1 standard deviation above the mean means a z-score of +1.
The easiest way to deal with z-scores is to sketch a curve (separate sketch for each problem)with the line of the mean down the middle. The mean has a z-score of 0. Each side of the mean has 50%, since the mean is in the exact middle of the normal curve. Draw a dashed line from the top of the curve straight down to the z-score you're working on. (Draw two lines if there are two z-scores in the problem.)
Look up the z-score in the z-score table. This gives you the % BETWEEN the mean and the z-score.
So for p(z>2.57), you'll subtract 50- the percentage for the z-score to get the percentage for the right tail.
For p(z<1.85) look up the % for 1.85. This is the % BETWEEN the mean and 1.85. But we've shaded the left half of the curve also. So we add the left 50% to the z-score % to get the total percent below the z-score line.
For p(-2.13<z<1.74, we've shaded a large middle area. So we have two lines. Work the right side and find the % BETWEEN the mean and 1.74. Then find the % between the mean and 2.13. Since they're on opposite sides of the mean line, ADD the percents.
Little sketches sure help visualize what's happening.
For p(-2.92<z<-1.46), find the percentages for the z-score as before. But this time we've shaded a much smaller area. Since these z-scores are on the same side of the mean, SUBTRACT the percents to find the percentage in the middle.
We want a right tail of 2.6%, so we know the z-core is +. Left tails are - z-scores.
We want just a sliver of the right end, so we subtract 50% (all of the right half) - 2.6% to find the percentage between the mean and z-score we're hunting for.
Then use the z-score table inside out. Find the percentage in the table and read the z-score for that percentage.
A left tail of 4.85% works just the same way, but the z-score is negative.
To calculate a z-score, subtract the value minus the mean, then divide by the standard dev.
37-56 = -19 --> 19/7= - 2.71
71 -56 = 15 --> 15/7 = ?
Now proceed as above.
Hope this is helpful. The key is to understand that in order to use the normal curve, we need to convert different means and standard deviations into the same units. (Just as comparing miles and kilometers requires conversion of units.) A z-score unit is simply 1 standard deviation away from the mean. Positive indicates above the mean and negative indicates below the mean.
This depends on the table you are using. Both Abramowitz and Stegun and
the Schaum Mathematical Handbook in their normal distribution tables give
the probability of a value less than the z-score. So these are greater than
you might expect by 0.5.
Also they in general do not give %'s but give straight probabilities in the
range 0-1.
RonL
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