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Thanks a lot, it really helped.The underlined phrase means the weight of 94% of the letters have a standard deviation of 12g. So let's say the mean is 20g. 94% of the letters have a weight of something between 18g and 42g, i.e. 12g off from the mean value.
As for the second question..
(i) Let the probability of the spinner landing on the blue side be 'x'. Since the probability of the spinner landing on the red and green sides are three and four times P(blue) respectively, their probabilities are '3x' and '4x'. Probabilities always add up to 1, so..
x + 3x + 4x = 1
8x = 1
x = 1/8
Hence the probability of the spinner landing on the blue side is 1/8.
(ii) You want a red AND a green AND a blue, so you multiply their probabilities:
1/8 * 4/8 * 3/8 = 12/512.
Also, since there are 3! different ways for this to happen (you could have RBG, BGR, GRB, etc.), you multiply the result by 3!
12/512 * 3! = 9/64.
(iii) Use normal approximation for this.
n = 136
p (success of landing on blue) = 1/8
q (failure of landing on blue) = 7/8
mean = 136 x 1/8 = 17
standard deviation = √(17 * 7/8) = √14.875
Also note that you will use continunity correction, so your value becomes 19.5, not 20. Using formula,
z = (19.5 - 17) / √14.875
Once you figure out the value of z, use the table to figure out the probability.
I hope this helped.
But I still don't get w11/p61/Q5.. The marking scheme has taken the probability of <32 to be equal to 0.97 in solution to part (i). How's that so?