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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
Learn more about Probability:
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