Area of a semicircle with radius of 6
2 answers:
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Solution :
Given data :
The mean length of the steel rod = 29 centimeter (cm)
The standard deviation of a normally distributed lengths of rods = 0.07 centimeter (cm)
a). We are required to find the proportion of rod that have a length of less than 28.9 centimeter (cm).
Therefore, P(x < 28.9) = P(z < (28.9-29) / 0.07)
= P(z < -1.42)
= 0.0778
b). Any rods which is shorter than cm or longer than
cm that re discarded.
Therefore,
P (x < 24.84 or 25.16 < x) = P( -59.42 < z or -54.85)
= 1.052