Hi, Welcome! Thank you for using the site. I'll be happy to help you with this, if you wish to proceed. What makes you think that you are on the wrong site? Ryan

Hi, Yes, I see a problem concerning "overhead reach distance of adult females", with parts A, B, and C. If there was more to it, or additional problems, then you will need to repost those parts. I'll have a solution for this problem posted for you shortly. Ryan

Hi, While it is possible to do that, it requires the use of the "Additional Services" feature, which adds a small additional charge. The smallest amount that the system will allow me to set it at is $5. I hate that this feels so very "arm-twisty", but this is the only way that the system will allow the exchange of personal information such as email addresses. If privacy concerning the solutions is not a concern, I can post the solutions directly to the page. They'll just be a little less "pretty" in plain text. Please let me know which option works better for you. Ryan

Hi, Here are the solutions: A. Find the probability that an individual distance is greater than 216cm. First, calculate the corresponding z-value: z = (x - µ) / sigma = (216 - 204)/8.5 = 1.4118 Then, the probability that an individual distance is greater than 216 cm is the same as the probability that a z-value will be greater than 1.4118. Using a statistical calculator gives: P(x > 216) = P(z > 1.4118) = 0.079 B. Find the probability that the mean for 23 randomly selected distance is greater than 202 cm. Again, first calculate the corresponding z-value: z = (xbar - µ) / (sigma / √n) = (202 - 204) / (8.5 / √23) = -1.1284 Then, the probability that a sample mean is greater than 202 is the same as the probability that a z-value will be greater than -1.1284: P(xbar > 202) = P(z > -1.1284) = 0.8704 C. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30? The normal distribution can be used for part (b) because the population standard deviation is known. If the population standard deviation is not known, and the sample size does not exceed 30, then the Student t distribution would be used instead of the normal distribution. Please don't hesitate to ask if you have any questions about these solutions. Thanks, Ryan

Ryan, the probability is the area to the right of z= 1.41 under the standard normal distribution curve, which I have shaded in the graph on the right, so then the area to the left is 0.9207? Correct? 1-0.9207= 0.0793Sorry I hate math, and it takes me longer than the average person to understand..:)

Hi, Yes, you have that correct. In part A, since we want the probability that x is GREATER than 216, we need the area to the RIGHT of z = 1.41. The exact procedure for determining that area depends on how your specific table is constructed. Some tables will give the area to the left of z, some will give the area between z = 0 and z, and some will give the tail area to the right of z. If you have the area to the left of z as 0.9207 from your table, then the area to the right of z will be 1 - 0.9207 = 0.0793. Ryan