Question

Consider the probability that no less than 88 out of 158 computers will not crash in a day. Assume the probability that a given computer will not crash in a day is 57%.

Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Solution:

Given that,

p = 57% = 0.57 ..probability of success of " will not crash"

1 - p = 1 - 0.57 = 0.43

n = 158

Let X be the number of computers in 158 that will not crash.

Here, BIN ( n , p ) that is , BIN (158 , 0.57)

Find P(X not less than 88) i.e. P(X 88)

n*p = 90.06

n(1- p) = 67.94

Since both are greater than 5 , we can use normal approximation to binomial.

According to normal approximation binomial,

X Normal

Mean = = n * p  =  90.06

Standard deviation = =[n*p*(1-p)] = [158 * 0.57 * 0.43] = 6.2230057

* Solution using continuity correction:

P(X 88) = P(X > 87.5) ..continuity correction factor

= P((x - ) / > (87.5 - 90.06) / 6.2230057)

= P(Z > -0.4114)

= 1 - P(Z <  -0.4114)

= 1 - 0.3404

= 0.6596

* Solution without using continuity correction:

P(X 88)  = P((x - ) / > (88 - 90.06) / 6.2230057)

= P(Z > -0.3310)

= 1 - P(Z <  -0.3310)

= 1 - 0.3703

= 0.6297