"\\bar{x}=2.112\\ lb, s=1.111\\ lb"
a. "x=5.64\\ lb"
b.
"\\dfrac{x-\\bar{x}}{s}=\\dfrac{3.528\\ lb}{1.111\\ lb}\\approx3.1755"The difference is approximately 3.1733 standard deviations.
c.
d.Since "3.1755>2," then the weight of 5.64 lb is significant.
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