Pleas Help Me, I Dont want the answer handed to me i just want some help starting... The mean number of matches in a particular size matchbox is 50 with a standard deviation of 2. If the number of matches per box is normally distributed, calculate the percentage of boxes with: (a) less than 47 matches (b) 49 or more matches.

Plz dont ignore it

I don't remember the formulas for normal distribution. You need to find how many matchboxes fall below 1.5 standard deviation and how many above 1/2 standard deviation.

you have \[X\sim N(50,2^2)\] and you want to calculate \[P(X<47)\] and \[P(X\ge49)\] use the fact that \[\frac{X-\mu}{\sigma}\sim N(0,1)\]

Formula : X < mean = 0.5-Z X > mean = 0.5+Z X = mean = 0.5 Z = (X-m) / σ where, m = Mean. σ = Standard Deviation. X = Normal Random Variable

Thanks All, Zarkon That is Perfect I Just Needed it In That Form To See How To Do It I Just Did A Load Just Like That

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