a)
The sample of 16 is normally distributed with the sample average 12 mm and
"\\sigma=4\/\\sqrt{16}=1" mm
b)
"z=\\frac{x-\\overline{x}}{\\sigma\/\\sqrt{n}}=\\frac{15.6-12}{4\/\\sqrt{16}}=3.6"
"P(z<3.6)=0.9998"
c)
"z_1=\\frac{12.8-12}{4\/\\sqrt{16}}=0.8"
"z_2=\\frac{13.2-12}{4\/\\sqrt{16}}=1.2"
"P(12.8<\\overline{x}<13.2)=P(z<1.2)-P(z<0.8)=0.0968"
Comments
Leave a comment