6. For a standard normal distribution (2), a. Fill in the blanks. b. P(Z > 1.24) astunim (8.11 (b) *2.5 (9) c. P(Z < 2.13) PAB (b) 30.0 (b) XEA.88 (b) d. P(-0.45 c) = 0.39. f. Find c if P(-c

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78.7 cm
from
15 to 18-
Chile
*2.5 (9)
c. P(Z < 2.13)
Week 7 worksheet
6. For a standard normal distribution (2), answer the following:
a. Fill in the blanks.
b. P(Z > 1.24) 29tunim fa.II (b)
$0.0 (9)
2= X-M
A8 (b)
30.0 (b)
XEA (b)
d. P(-0.45<z < 2.29)
e. Find c if P(Z > c) = 0.39.
4
f. Find c if P(-c<z< c) = 0.83.
x=141+2(0)
Smol
Name:
anor BE.IS equor A8A (5)
270
88.0 (5)
ACV (d)
81.0 (d)
28.CA () AT bris 08 (d)
ea.o (p)
[1.0 (d)
0 (d)
A8 IT (a)
tsolzshow TossW
Uslh69
the
T
wo Hoy no 00 (r)
Z~ N M G
nwo muoy no on oll
02 oll
nwo way no 00 A
nwo nupy no oQ (s) 2
nwo Twoy no cd (98
af.0 (n) T
area bns b8 30 (1)
ماما .
.28
1.37
The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of
three. Let X = golf score for a random school team, then fill in the blanks:
a. Find the probability that a randomly selected golfer scored at most 65.
Then X~ N 6₂8 3
_).
Transcribed Image Text:78.7 cm from 15 to 18- Chile *2.5 (9) c. P(Z < 2.13) Week 7 worksheet 6. For a standard normal distribution (2), answer the following: a. Fill in the blanks. b. P(Z > 1.24) 29tunim fa.II (b) $0.0 (9) 2= X-M A8 (b) 30.0 (b) XEA (b) d. P(-0.45<z < 2.29) e. Find c if P(Z > c) = 0.39. 4 f. Find c if P(-c<z< c) = 0.83. x=141+2(0) Smol Name: anor BE.IS equor A8A (5) 270 88.0 (5) ACV (d) 81.0 (d) 28.CA () AT bris 08 (d) ea.o (p) [1.0 (d) 0 (d) A8 IT (a) tsolzshow TossW Uslh69 the T wo Hoy no 00 (r) Z~ N M G nwo muoy no on oll 02 oll nwo way no 00 A nwo nupy no oQ (s) 2 nwo Twoy no cd (98 af.0 (n) T area bns b8 30 (1) ماما . .28 1.37 The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Let X = golf score for a random school team, then fill in the blanks: a. Find the probability that a randomly selected golfer scored at most 65. Then X~ N 6₂8 3 _).
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