Question

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Exit Suppose that car wash times are normally distributed with a mean of 25 minutes and a standard deviation of 3 minutes. You will be considering a group of 9 randomly selected car wash times. Answer the following questions. D 1. Given the information above, what can we say about the sampling distribution of x-bar. It is also normally distributed, because the original poplation was normally distributed. It is normally distributed because n is bigger than 30. We do not know since n is not at least 30. D 2. Given the information above, what are the mean and st. dev of x-bar? mean = 25 st dev=1 mean = 25 st dev = 3 mean = 25 st dev = 9

Answer 1

= as noq ou xnn (25, (3/²) As population is normally distributed Sample mean will also be normally meln norm distributed eption A llp= l = 35 st deviation - 1 mean = 25 eption A Pliu 326) = P( 28 26-25 ) plă 326) = 1- P(241) Plüsao) = 1- 0.8413 Cusing PC320) = 0.158+) I Z-Table

P( 23 <ła 27.5) =P(u <37-5) - P(u L23) -|(2117.5-25 )- P(Ž 211-25 I ) I al(229.75) - P(22-2) a -tible P(228.7512 0.9970 P(22-) = 0.0228 using P(23252 227,5) = 0.9970 – 0.0927 Pas zā<37.5) - (0.97427 P(23 x 97.5) 0.9742