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Unformatted text preview: 6.8 6.25 “Fe want to find a :\7alue such that Edi—:0 < : < :0 ) = .8262 (see below). z'n [i 21': Since Aka 1—1 (—:0
that the lower tail area must be Ali—:0 ] = .123 8/2 = .0869 . From Table 3. —:ﬂ = —1.36 and :S 21.36. } = .8262 ._ the total area in the two tails of the distribution must be 1—.82 62 = .1238 so It is given that :t. the tuisupported stem diameter of a sunﬂower plant. is normally distributed with ,1! = 35 2111:1023. _ s
a P(.1'>40}=P' :::= 40 3"
I. 3, j 1) From part a. the probability that one plant has stem diameter of more than 40 mm is .0425. Since the
two plants are independent. the probability that two plants both have diameters of more than 40 mm is [i .0425 .0425} = .00226 : Since 95% of all 111easurements for a normal random variable lie within 1.96 standard deviations of the
mean. the necessary interval is =P[: :} 1.6?) = 1—.9525 = .0425 11:1.965 :> 35 i1.96(3) :> 35 i 5.88 or in the interval 29.12 to 40.88.
(1 The 90r11 percentile of the standard normal distribution was found in Exercise 6.11a to be : = 1.28 .
Since the relationship between the general normal random variable x and the standard normal: is I — H . . . . . .
: = ‘ . the corresponding percentile for this general normal random Tvariable is found by solvmg for 0' x = '11 + :o‘ . I = 35+].28(3) or .r = 38.84 6.30 It is given that :t' is normally distributed with ,H unknown and 0' = 25.? . It is necessary to have P l: x > 2000] = .01. Calculate 2000'— .f . 2000'— a”
:=—‘£ and P(1‘;~2000]=P:>—'ﬁ =
25.? \ 25.7 , Since the value (2000 :1: U252 is a constant (although its value is unknown). it can be treated as such in .0] Table 5. It is necessaiy to ﬁnd :0 such that P[: ::: :CI } = .01 01‘ A[:D:}=.9900.
From Table 3. :0 = 2.33 . so that 2000 — ,u _ —2.33 and ,6! 21940.119 2 5. ...
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 Spring '10
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