If both of np and nq are greater than or equal to 10, then the sampling distribution of sample proportions will be approximately normal with mean p and variance "\\frac{pq}{n}" .
"\\hat{p}~ N(P, \\frac{pq}{n})"
p = population proportion,
q = 1 - p
n = sample size
p = 0.55
q = (1 - 0.55) = 0.45
n = 500
np = 275, which is greater than 10.
nq = 225, which is greater than 10.
Hence sampling distribution of sample proportions will be approximately normal.
We have to find P("\\hat{p} < 0.49" ).
"Z = \\frac{\\hat{p}-p}{\\sqrt{\\frac{pq}{n}}} \\\\\n\nP(\\hat{p}<0.49) = P( \\frac{\\hat{p}-p}{\\sqrt{\\frac{pq}{n}}} < \\frac{0.49-p}{\\sqrt{ \\frac{pq}{n} }}) \\\\\n\n= P(Z< \\frac{0.49-0.55}{\\sqrt{ \\frac{0.55 \\times 0.45}{500} }}) \\\\\n\n= P(Z< -2.696) \\\\\n\n= 0.035"
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