## Answers

TOPIC:Normal distribution and probability.

Suppose (x~ N far, 60) @ we need, (4-26 GX 2 M +20) = P ( 1.20-me (X-M ( ut 20 m) =P(-2 (242). [2 = x u-20-M X - M . Mt 25-M NNI = 8 (2) – 8 (-2). [ Dcy = cdf of 2). = 9(2) - {1-012).

Tu C-k) = 1DU) = 2 D (2) - 1. = 2 x 0.9772-1 [ From Normal Itable values. 0.9544 Ans]

6 we need, P (M-1.50 cх(M+ 2.50) =2(1500ML CRL < 1420505-M). = P(-1.5 42 42.5), = D. (2.5) – 8 (-1.5).

.
= 0(3:5) - {1-7 (1.5)) [from Here - 938- 30.99 - 10. 9270] [Ane] I from Normall - 0.9938 table We need, && (x2m-30). =p (Rage < Me-36-) = 8 (2 4-3) = (-3). = 1- P13). = 1- 0.9987.

- 0.0013 Doma rounded to four Answers are decimals)

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