1 Let Z N0 1 nd value c such that Pc Z c 06922 2 Let Y N8

1. Let Z N(0; 1), nd value c such that P(??c < Z < c) = 0.6922.
2. Let Y N(??8; 4), calculate P(Y < ??5) and P(??2 < Y < 7).
3. Given sample data (1.3,2.5,2.4,3.5,6.1,2.2) from normal population, cal-
culate sample median, sample mean and sample variance.
4. Let X 2
13, nd critical value 2 0:025;13 and 2 0:975;13.
5. Find critical value F0:05(5; 6) and F0:05(4; 10) for F-distribution.

Solution

(1) P(-c<Z<c) = 0.6922

--> 2*P(0<Z<c) =0.6922

--> P(0<Z<c) = 0.6922/2 =0.3461

--> P(Z<c) - P(Z<0) =0.3461

--> P(Z<c) -0.5= 0.3461

--> P(Z<c) = 0.3461+0.5=0.8461

So c= 1.02 (from standard normal table)

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(2)

P(Y<-5) = P((Y-mean)/s <(-5+8)/2) = P(Z<1.5) = 0.9332 (from standard normal table)

P(-2<Y<7) = P((-2+8)/2 <Z< (7+8)/2)

=P(3<Z<7.5) = 0.0013 (from standard normal table)

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(3) sample median= 2.45

sample mean= 3

sample variance= 2.8

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(4)

X^2 with 0.025, 13 = 5.01 (from chisquare table)

X^2 with 0.975, 13 = 24.74 (from chisquare table)

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(5)

F0.05 (5,6) = 0.20 (from F table)

F0.05(4,10) = 0.17 (from F table)