Question

Random samples of size *n* = 90 were selected from a
binomial population with *p* = 0.3. Use the normal
distribution to approximate the following probability. (Round your
answer to four decimal places.)

* P*(0.26 ≤

Answer #1

Using normal approximation,

P( < p ) = P( Z < ( - p) / sqrt [ p ( 1 - p) / n ]

So,

P( 0.26 < < 0.34) = P( < 0.34) - P( < 0.26)

= P(Z < ( 0.34 - 0.30) / sqrt [ 0.30 (1 - 0.30) / 90] < p < P(Z < ( 0.26 - 0.30) / sqrt [ 0.30 (1 - 0.30) / 90]

= P(Z < 0.83) - P(Z < -0.83)

= 0.7967 - 0.2033 (From Z table)

= **0.5934**

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