A=
AP=<t><c></t>
C0=
C1=
C2=
C3=
C4=
C5=
C6=
C7=
C8=
C9=
ES=Answers may vary.
F0=
F1=V13
F2=V7
H=$.hint$
INST=
M1=
M2=
M3=
M4=
M5=
M6=
M7=
M8=
M9=
MS=1
MW1=
MW2=
MW3=
MW4=
MW5=
MW6=
MW7=
MW8=
MW9=
N=
Q=Part of a test given to young children consists of\nputting together a simple jigsaw<br> puzzle. Suppose\nthat such a puzzle is given to V14 children, each\nchild is timed, and<br>a graph of the times is made. Suppose that the graph is a normal curve with a\nmean<br>time of V8 seconds and a standard deviation\nof V9 seconds.<br>V4<br>$.tablea1$
SA=Answers may vary.
T=F
TF=-1
TL=
TOL=+1E-4
U=NOUNIT
V0=I[1,3]
V1=S"About how many of the children finished the puzzle in less than V12 seconds?"
V10=I(2*V9)
V11=I(V8+V10)
V12=I(V8-V10)
V13=L[V0:V15,V15,V16]
V14=I[1000,2000,100]
V15=I(V14*0.25)
V16=I(V14*0.68)
V17=L[V0:V12,V11]
V18=L[V0:left, right]
V19=L[V0:less,more]
V2=S"How many took more than V11 seconds?"
V3=S"If you rated as "average" all the children within 1 standard deviation from the mean, how many children would fall into this classification?"
V4=L[V0:V1,V2,V3]
V5=S"$mu$"
V6=S"$sigma$"
V7=L[V0:students,students,students]
V8=I[100,200,10]
V9=I[11,19]
W=<step>In this problem, we are given that\nV5 = <box>V8</box> sec and V6 = V9 sec.\n</step>\n<ostep>V0!=3::\nThe number V17 is 2 standard deviation from the mean V8 and 2.5% of students are to the V18 of V17. \n2.5% of V14 = <box>V15</box> students took V19 than V17 seconds to complete the puzzle.\n</ostep>\n<ostep>V0==3::\n \nAs you can see from the diagram, 68% of the children are within one standard deviation of the<br> mean and 68% of V14 is <box>V16</box>. Thus, there are <box>V16</box> children within one standard deviation from<br> the mean.\n</ostep>\n
YN=-1
cnum=4
followup=
ilev=0
mcdm=1
mdm=0
mpc=unspecified
mpn=2
mpv=2.0
mwnum=2
ncd=-1
subject=unspecified
varnum=24
commentV8=        //mu
commentV9=        //sigma