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The salary of an electrician is normally distributed with mean £{m} and standard deviation £{s}.

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(a)

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Converting to $\\operatorname{N}(0,1)$

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$\\simplify[all,!collectNumbers]{P(X > {upper}) = P(Z > ({upper} -{m}) / {s})} = P(Z>\\var{zupper}) = 1-P(Z<\\var{zupper})=1-\\var{p1} = \\var{prob2}$ to 2 decimal places.

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(b)

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$\\simplify[all,!collectNumbers]{P({lower} < X < {upper}) = P(X < {upper})-P(X < {lower})}=P(Z<\\var{zupper})-P(Z<-\\var{zlower}) =\\var{p1}-\\var{p2} = \\var{prob3}$ to 2 decimal places.

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\n

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Given a random variable $X$  normally distributed as $\\operatorname{N}(m,\\sigma^2)$ find probabilities $P(X \\gt a),\\; a \\gt m;\\;\\;P(X \\lt b),\\;b \\lt m$.

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rebelmaths

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If one electrician is chosen at random, what is probability that this electrician earns over £{upper}? 

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Probability = [[0]](to 2  decimal places)

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If one electrician is chosen at random, what is probability that this electrician earns between £{lower} and £{upper}?

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Probability = [[0]](to 2  decimal places)

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