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Sagot :
We would test if np and n(1 - p) is at least 10
p = sample proportion
p = x/n
where
x = number of success
n = sample size
From the information given,
p = 41% = 41/100 = 0.41
n = 118
np = 48.38
n(1 - p) = 118(1 - 0.41) = 69.62
Both values are greater than or equal to 10. Thus, the distribution is approximately normal.
The mean of the distribution = p = 0.41
Standard deviation = √(p(1 - p)/n
By substituting the values,
standard deviation = √(0.41(1 - 0.41)/118 = √0.00205 = 0.045
We want to find P(p ≥ 0.5)
We would standardize 0.5 into a z score by applying the formula,
z = (x - mean)/standard deviation
z = (0.5 - 0.41)/0.045 = 2
From the normal distribution table, the probability value corresponding to a z score of 2 is 0.9773
P(p ≥ 0.5) = 1 - 0.9773
P(p ≥ 0.5) = 0.023
the probability that the majority (more than 50%) watched the Super Bow is 0.023
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