Question

Let x denoted the number of flaws along a 100 m reel of magnetic tape (values of x are whole numbers). Suppose x has approximately a normal distribution with u=25 and 0=5. a. What proportion of reels will have between 20 and 40 flaws, inclusive? (Ans:0.8633) b. What proportion of reels will have at most 30 flaws? Fewer than 30 flaws? (Ans:0.8643, 0.8159)

Answer 1

Sok given that the data i * Let x denoted the number of flaws along loom reel of magnetic tape * given nolmal distribution with el=25 and a=5 a) the finding proportion between of reels will have 20 and 40 flaws, inclusive continuity PC 2o2xCHO)- P (20-0.5 <x<40 + = P(19.52 X <dy 0.5) excitp +0.5) Los more Lion)

☺ Ep(19.5=905 5 x=25€ 40.525) =P(-11E X-M = 3.1) = P(-111 2 2 4 3.1) = 0.8633 P(Z <-11) = 1-plz cli) - 1-0.8643 = 0.1357 P(Z <3. 1) = 0.999 0.999-0.1357 PERIC 223.1) = 0.8633 b) Now proportion at most 30 flaws of reek will have and fewer han 30 flaws i at most 30 flaws 7) P(x430) =P(XM £ 30.5-25) = P(ZE!) = 0.8643 (1) fewer than 30 flaws - P(x230) 30.5-25 = P(X-M a 29.5-25 = P(Z 21.1) = 0.8159