Question

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. A symmetric bell-shaped curve is plotted over a horizontal scale with two labeled coordinates. One coordinate is labeled "0" and is located at the center and peak of the curve. The other coordinate is labeled "z," and is to the left of 0. A vertical line extends from the scale to the curve at z. The area under the curve to the right of z is shaded and labeled “0.7611.” The indicated z score is nothing. The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.3 cm. a. Find the probability that an individual distance is greater than 215.5 cm. b. Find the probability that the mean for 15 randomly selected distances is greater than 204. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? a. The probability is nothing. (Round to four decimal places as needed.) b. The probability is nothing. (Round to four decimal places as needed.) c. Choose the correct answer below. A. The normal distribution can be used because the finite population correction factor is small. B. The normal distribution can be used because the original population has a normal distribution. C. The normal distribution can be used because the mean is large. D. The normal distribution can be used because the probability is less than 0.5

          Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. A symmetric bell-shaped curve is plotted over a horizontal scale with two labeled coordinates. One coordinate is labeled "0" and is located at the center and peak of the curve. The other coordinate is labeled "z," and is to the left of 0. A vertical line extends from the scale to the curve at z. The area under the curve to the right of z is shaded and labeled “0.7611.” The indicated z score is nothing.

The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.3 cm. 

a. Find the probability that an individual distance is greater than 215.5 cm. 
b. Find the probability that the mean for 15 randomly selected distances is greater than 204.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

a. The probability is nothing. (Round to four decimal places as needed.)
b. The probability is nothing. (Round to four decimal places as needed.)
c. Choose the correct answer below.
A. The normal distribution can be used because the finite population correction factor is small.
B. The normal distribution can be used because the original population has a normal distribution.
C. The normal distribution can be used because the mean is large.
D. The normal distribution can be used because the probability is less than 0.5
        
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Added by Scott M.

Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. A symmetric bell-shaped curve is plotted over a horizontal scale with two labeled coordinates. One coordinate is labeled "0" and is located at the center and peak of the curve. The other coordinate is labeled "z," and is to the left of 0. A vertical line extends from the scale to the curve at z. The area under the curve to the right of z is shaded and labeled “0.7611.” The indicated z score is nothing. The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.3 cm. a. Find the probability that an individual distance is greater than 215.5 cm. b. Find the probability that the mean for 15 randomly selected distances is greater than 204. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? a. The probability is nothing. (Round to four decimal places as needed.) b. The probability is nothing. (Round to four decimal places as needed.) c. Choose the correct answer below. A. The normal distribution can be used because the finite population correction factor is small. B. The normal distribution can be used because the original population has a normal distribution. C. The normal distribution can be used because the mean is large. D. The normal distribution can be used because the probability is less than 0.5
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Transcript

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00:01 So your first question is dealing with finding out what the z value is, and i believe this is what your picture is supposed to look like, that you have the standard normal distribution with zero being in the center, and you have a z value that is to the left, and the area to the right of that is 0 .7611.
00:19 And so you can look up in your table of areas if you want, 0 .767676.
00:32 And find out that that corresponds to a z value of .71 and sorry when i write while i'm recording it does not identify my stylist very well in any case but we know that this z value is negative and you can also subtract and find the complement down here so that you can look it up directly but the z value is negative .71 now i'm on the next problem, you have a normal distribution with this mean and this standard deviation.
01:08 And you want to find the likelihood and you're told it's normal.
01:11 And so you want to find what the likelihood is of having a jump or distance that is 215.
01:18 So we need to convert that to a z value.
01:21 So we take this score minus the mean divided by the standard deviation and that gives us a z score rounded off to two decimals of 1 .2.
01:30 Now, once again, you can look up in your table 1 .2, but it's actually easier to look up symmetrically the area below negative 1 .2 because that will be the same is the area above positive 1 .2 and get that answer directly without subtracting.
01:49 So if you just look at a picture, it makes the symmetry easier.
01:52 Now, they tell us now that we're getting a measurement of 15 people in finding a mean, and we want to know what's the likelihood.
02:00 That the mean is higher than 204...
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