6fbd41904fbOe 39143747a452fc23d7ds 4852880 Take a Test-Arian Khorsandi - Google
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6fbd41904fbOe 39143747a452fc23d7ds 4852880 Take a Test-Arian Khorsandi - Google Chrome ttps://www.mathod.com/Student/Player Test.aspx?testld-1319103728centerwinsyes&fromPlayerCheck; yes Arian Khorsandi 4/23/16 7:s0 lz: Chapter 6 Review Quiz Overvie 0 0 0 0 0 0 0 0 0 " > This Question 3 et this Test Tame Remaining: 01 09 35 In a stady on the physical acthity of young adults, pediatric researchers measared overall physical activity as the total aamber of registered overal physical activity of obese young aduits has a mean of u-320 cpem and a standard deviation of a 90 cpm In a random sample of n 100 obese young adults, consider the sample mean counts per minute, . D Click the icen to view the table of sermal carve arcas a. Describe the smping ástrbtion oft Finds Find o o-(Type an ineger or a decimal) Descrbe the shape of the sampling distrbution ofx OA Apprcocmately normal OB Skewed right Oc Skewed let b. What is the probablity that the mean overall physical activity level of the sample ia between 300 and 310 ce PO00 sis 310)Round to four decimal laces as needed) c. What is the probabiliny that the mean overill physical activity level of the sample is greater than 368 cpm? PG2364)-(Roud to four decimal places as terded) Previous QuestionNeat GwestonExplanation / Answer
The mean, mux=320
Standard deviation, sigma/sqrt n=90/sqrt 100=9
Approximately normal. [per Central Limit Theorem]
From information, Xbar=320, s=9, find Z scores corresponding to Xi 300, 310
Z1=(Xi-Xbar)/s=(300-320)/9=-2.22 and Z2=(310-320)/9=-1.11
P(300<X<310)=0.4868-0.3665=0.1203
P(X>368)=P[Z>(368-320)/9]=P[Z>5.33]=0.4999
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