In your class work groups (the groups you have been working in over the last few classes) - complete the following homework assignment and post as a group:
Elementary Statistics⎯Picturing the World:
1. Section 5.1, Exercise #26, p. 225
2. Section 5.1, Exercise #60, p. 228
3. Section 5.2, Exercise #12, p. 232
4. Section 5.3, Exercise #44, p. 244
5. Section 5.4, Exercise #18, p. 255; do not sketch
the graph
Subscribe to:
Post Comments (Atom)
Tonya
ReplyDeleteRick Oneil
#26 - .0401
#60 - .95
#44 - A = 18.2
B = 19.75
#18 - 3.42
Jace, Alex, Justin
ReplyDelete26. .96
60. .9750 -.0250=.95 or 95% if the area is between -1.96 and 1.96.
12. .9535-.0548=.4055 or 41%
44. a) -.67 .67 so z=.67 = u16.5/o2.5 so .67=x-16.5/2.5 x=18.175lbs p.239
b.) -1 z=-1.03 = u16.5/o2.5 so -1.03=x-
16.5/16.5 x=13.925lbs
18. mean=49.3 sd=17.1, 17.1/sqrt25=3.42
51 +-3= 48->through54, and 49+-3=46->52.
Evan Dossey, Rick Robinson, Mel Bailey
ReplyDelete1. Section 5.1, Exercise #26, p. 225
1 - 0.0401 = 0.9599
2. Section 5.1, Exercise #60, p. 228
1 - (0.025 x 2) = .95
3. Section 5.2, Exercise #12, p. 232
p=.42
4. Section 5.3, Exercise #44, p. 244
A.) z=(x-16.5)/2.5 = 2.5x.68+16.5=x=18.2
B.) z=(x-16.5)/2.5 = 2.5x -1.3+16.5 = 12.75
5. Section 5.4, Exercise #18, p. 255; do not sketch
the graph
s=17.1/ sr25 = 17.1 / 5 = 3.42
Posted By: Bryan S., Sheila W., Karolyn M., Nick N.
ReplyDelete26) 0.09599 or 95%
60) 0.95
12) Z=-1.44
44) A. X=18.175
B. X=13.925
18) Mean is 49.3
Standard error of mean is 3.42
Tonya
ReplyDeleteRick Oneil
Correction to #44 B. 12.75
1. .0401
ReplyDelete2. 0.95
3. 43%
4.
A) 18.19 LBS
B) 13.91 LBS
5. 3.42