study by a class
x= single trout
mean is = 10.5 inches
standard dev is= 1.4 inches
probability a single trout is between 8 and 12 inches?
probability that length of 5 trout at random is between 8 and 12 inches?
the researchers are 93.8% confident
is there enough evident to support the research?
construct 93.8% confidence interval about the population mean.
construct a 92.28% confidence interval about population mean.
Z = (score-mean)/SD (one trout)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores.
Z = (score-mean)/SEm
SEm = SD/√n
Use Same Table.
93.8% = mean ± Z (SD)
Look up Z = .031 in table (100 - .938 = .062, half is .031)
Do same for 92.8%.