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You need to remember the formula of probability in this case of normal distribution such that:P(X>Y)
X denotes the mean of sample of 3
Hence `P(XgtY) = P(Xgt115)`
The mean of sample of 3 needs to be larger than `(X-Y)/sqrt(15^2/3)`
Notice that the standard distribution expresses the square root of variance , hence `P(Xgt=(115-100)/sqrt(15^2/3)) =gt P(Xgt= 2.23) = 0.04`
Hence, since the chances for the mean of a sample of 3 to be 115 are to small, about 4 %, then it would be unusual to happen.
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