tomer. Find the probability tHal ils to meet customer specifications. 5-60 The w

Question

tomer. Find the probability tHal ils to meet customer specifications. 5-60 The width of a casing for a door is normally dis tributed a with a mean of 61 cm and a standard deviation of 0.4 cm. The width of a door is normally distributed with a mean of 60.5 cm and a standard deviation of 0.2 cm. Assume independence. (a) Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. (b) What is the probability that the width of the casing minus the width of the door exceeds 0.8 cm? (c) What is the probability that the door does not fit in the casing?

Explanation / Answer

X = width of casing , X ~ N(61,0.4^2)
Y = width of door , Y ~ N(60.5,0.2^2)
a) mean of X - Y = 61-60.5 = 0.5
sd of (X - Y) = sqrt(0.4^2 + 0.2^2) = sqrt(0.20)
b)
P(X - Y > 0.8)
= P(Z> (0.8 - 0.5)/sqrt(0.20))
=P(Z> 0.6708203)
= 0.2512

c)
P(X - Y < 0)
= = P(Z> (0 - 0.5)/sqrt(0.20))
=P(Z> -1.118033)
= 0.8682

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