*X* = length of part

(a) `20.03` is `1` standard deviation below the mean;

`20.08` is `(20.08-20.05)/0.02=1.5` standard deviations above the mean.

`P(20.03 < X < 20.08)`

`=P(-1<Z<1.5)`

`=0.3413+0.4332`

`=0.7745`

So the probability is `0.7745`.

(b) `20.06` is `0.5` standard deviations above the mean;

`20.07` is `1` standard deviation above the mean

`P(20.06 < X < 20.07)`

`= P(0.5 < Z < 1)`

`=0.3413-0.1915`

`=0.1498`

So the probability is `0.1498`.

(c) `20.01` is `2` s.d. (standard deviations) below the mean.

`P(X<20.01)`

`=P(Z < -2)`

`=0.5-0.4792`

`=0.0228`

So the probability is `0.0228`.

(d) `20.09` is `2` s.d. above the mean, so the answer will be the same as (c),

`P(X > 20.09) = 0.0228.`

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