The iq of 300 students in a certain school in Quezon province is approximately normally distributed with mean of 100 and standard deviation is 15??
Sample size (n) = 300
Mean = 100
Standard deviation = 15
The percentage of students have an IQ from 85 to 120 is:
"P(85<X<120)=P(\\frac{X-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}<Z<\\frac{X-\\mu}{\\frac{\\sigma}{\\sqrt{n}}})"
"P(85<X<120)=P(\\frac{85-100}{\\frac{15}{\\sqrt{300}}}<Z<\\frac{120-100}{\\frac{15}{\\sqrt{300}}})"
"P(85<X<120)=P(-17.32<Z<23.09)"
"(85<X<120)=P(Z<23.09)-(Z<-17.32)"
Find probability in respect to each Z score using standard distribution table.
"(85<X<120)=1.0-0.00=1.0\\approx100\\%"
This means approximately 100% students have an IQ from 85 to 120.
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