Let p be
the population proportion. <span>
We have p=0.60, n=200 and we are asked to find
P(^p<0.58). </span>
The thumb of the rule is since n*p = 200*0.60
and n*(1-p)= 200*(1-0.60) = 80 are both at least greater than 5, then n is
considered to be large and hence the sampling distribution of sample
proportion-^p will follow the z standard normal distribution. Hence this
sampling distribution will have the mean of all sample proportions- U^p = p =
0.60 and the standard deviation of all sample proportions- δ^p = √[p*(1-p)/n] =
√[0.60*(1-0.60)/200] = √0.0012.
So, the probability that the sample proportion
is less than 0.58
= P(^p<0.58)
= P{[(^p-U^p)/√[p*(1-p)/n]<[(0.58-0.60)/√0...
= P(z<-0.58)
= P(z<0) - P(-0.58<z<0)
= 0.5 - 0.2190
= 0.281
<span>So, there is 0.281 or 28.1% probability that the
sample proportion is less than 0.58. </span>
For the equation of the line of the form y = mx + b, m is the slope of the line and b is the y-intercept which is the value of y when x is equal to zero. The slope is the rate of change of y per change in x. "y" would represent the dependent variable which is the weight of the baby while "x" represents the independent variable which is the number of months or the age of the baby in months. We calculate the slope as follows:
slope = (11 - 9) / ( 4 - 0) = 2/4 = 0.5
at x = 0, it is said that the weight of the baby is 9 lbs so the value of b would be 9.
The equation of the line would be y = 0.5x + 9
6x^-12 - 6y^3 ... I think
