概率统计初步In USA,2% of the 2 million students who take the SAT each year receive special accommodations because of disasbilities.Consider a random sample of 25 students who have recently taken the test.What's the probability that the number a
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![概率统计初步In USA,2% of the 2 million students who take the SAT each year receive special accommodations because of disasbilities.Consider a random sample of 25 students who have recently taken the test.What's the probability that the number a](/uploads/image/z/8664623-71-3.jpg?t=%E6%A6%82%E7%8E%87%E7%BB%9F%E8%AE%A1%E5%88%9D%E6%AD%A5In+USA%2C2%25+of+the+2+million+students+who+take+the+SAT+each+year+receive+special+accommodations+because+of+disasbilities.Consider+a+random+sample+of+25+students+who+have+recently+taken+the+test.What%27s+the+probability+that+the+number+a)
概率统计初步In USA,2% of the 2 million students who take the SAT each year receive special accommodations because of disasbilities.Consider a random sample of 25 students who have recently taken the test.What's the probability that the number a
概率统计初步
In USA,2% of the 2 million students who take the SAT each year receive special accommodations because of disasbilities.Consider a random sample of 25 students who have recently taken the test.What's the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?
概率统计初步In USA,2% of the 2 million students who take the SAT each year receive special accommodations because of disasbilities.Consider a random sample of 25 students who have recently taken the test.What's the probability that the number a
think about 0-1 distribution,
n = 25,p = 0.02.
np = 0.5,
np(1-p) = 0.5*(1-0.02) = 0.49.
X can be treated as normal distribution N(0.5,0.49)
Y = (X - 0.5)/0.7 N(0,1)
then,
the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated
= P{|Y|