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a college report states that 30% of its students commute to school. if a random sample of 550 students is taken, find the probability that at least 143 commute. round your percentage to the hundredths place.

Sagot :

The probability of at least 143 commuting from a random sample of 550 students is 0.9796 or 97.96%

Probability of students who commute to school p = 0.30

Number of students n = 550

Using normal distribution approximation to the binomial distribution:

Mean µ = np = 550*0.30 = 165

Standard deviation σ = √np(1 -p) = √550*0.30(1 – 0.3) = 10.75

The require probability is P(X > 143)

Finding the z score of X,

z = (X - µ)/σ

z = (143 - 165)/10.75

z = -2.0465

Hence,

P(X > 143) = P(z > -2.0465) = 1 – P(z ≤ - 2.0465)

From the table,

P(z ≤ - 2.0465) = 0.020354

Hence,

P(X > 143) = 1 – 0.020354 = 0.979

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