Can someone please explain how to do this question, because I have 7 others just like it?

first, you must find the median and standard deviation of the amount fo green candies inside the bag.

median: (.15)(200) = 30 this should be fairly obvious.

standard deviation: there's a property you have to memorize to find this. its: sd = square root( (n) (p) (1-p)). in this case, its square root( (200) (.15) (.85)). there's no obvious reason for this, you just have to memorize. the value calculates to about 5.

now, you can calculate the z score. its median + (z)(standard deviation) = proposed value.

in this case, its 30 + 5z = 25. using simple algebra, you calculate z to be -1.

now you just look up -1=z on a normal cdf chart or use the NCDF function on your Calculator.

NCDF calculates the probability of the value falling on or before the z score. since you want less than 25 green candies (less than -1 z score) the CDF works. if you wanted greater than 25 candies, you would want to calculate 1 - the -1z CDF value.

http://www.uvm.edu/~dhowell/StatPages/More_Stuff/n...

this is a CDF calulator. the p value is your probabilty.

The probability that n pieces are green and 200-n pieces are not is

p(n) = 200Cn (0.15^n)(0.85^(200-n))

where 200Cn = 200! / (n! (200-n)!)

The probability that a bag contains less than 25 green candies would therefore be

p(0) + p(1) + ... + p(24)

According to Mathematica, this is 13.68%

The de Moivre–Laplace theorem says that a binomial distribution of a sample of size N and probability of success p can be approximated by a normal distribution with mean Np and variance Np(1-p).

In this case X = B[200, 0.15] can be approximated by Z = N[30, 25.5]. We want to approximate P(X < 25). It is standard to make a "continuity correction" and change this to P(X < 24.5). We compute Z = (24.5 - 30) / sqrt(25.5) = -1.0891622973.

so P(X < 24.5) = P(Z < -1.09) = 1 - P(Z < 1.09) = 1 - 0.86214 = .1389 = 13.79%