Answer to Question #320031 in Statistics and Probability for Mirylle Anne

Question #320031

A process yields 10% defective items. If 100 items are randomly selected the process, use normal approximation to find the probability that the number of defectives.

a) exceeds 13 and

b) less than 8.


1
Expert's answer
2022-04-01T04:55:08-0400

Let X be the number of defectives in the 100 items.

Use the Normal Approximation to the Binomial Distribution Theorem.

For large 𝑛, 𝑋 has approximately a normal distribution with Β΅ = 𝑛𝑝 and

𝜎

2 = 𝑛𝑝(1 βˆ’ 𝑝) and

P(X<x)=P(z<xβˆ’0.5βˆ’npnp(1βˆ’p))P(X<x)=P(z<\frac{x-0.5-np}{\sqrt{np(1-p)}})

P(X≀x)=P(z<x+0.5βˆ’npnp(1βˆ’p))P(X\le x)=P(z<\frac{x+0.5-np}{\sqrt{np(1-p)}})

P(X>x)=P(z>x+0.5βˆ’npnp(1βˆ’p))P(X>x)=P(z>\frac{x+0.5-np}{\sqrt{np(1-p)}})

P(Xβ‰₯x)=P(z>xβˆ’0.5βˆ’npnp(1βˆ’p))P(X\ge x)=P(z>\frac{x-0.5-np}{\sqrt{np(1-p)}})

We have that

𝑝 = 0.1, 𝑛 = 100, 𝑛𝑝 = 100(0.1) = 10, np(1βˆ’p)=100(0.1)(1βˆ’0.1)=3\sqrt{np(1-p)}=\sqrt{100(0.1)(1-0.1)}=3

a. P(X>13)=1βˆ’P(X≀13)=1βˆ’P(13+0.5βˆ’103)=1βˆ’P(Z≀1.17)=0.5βˆ’0.3790=0.121P(X>13)=1-P(X\le13)=1-P(\frac{13+0.5-10}{3})=1-P(Z \le 1.17)=0.5-0.3790=0.121

b. P(X<8)=P(X<8βˆ’0.5βˆ’103)=P(Z<βˆ’0.83)=0.2033P(X<8)=P(X<\frac{8-0.5-10}{3})=P(Z<-0.83)=0.2033







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