5.5 The inside diameter, X, in millimeters of a randomly selected nozzle is norm

Question

5.5 The inside diameter, X, in millimeters of a randomly selected nozzle is normally distributed with mean 11 and standard deviation 0.25.

(a) Sketch the pdf of X, shade the area that represents P[X 11.5] and find this probability.

(b) Compute P[11.1 X 11.5].

(c) Find a diameter, x sub 0, so that exactly 5% the diameters of such nozzles exceed x sub 0.

5.6 The temperature, T, in degrees Celsius after a chemical reaction is normally distributed with mean 50 and variance 4.

(a) Compute P[48 < T < 53.5].

(b) Compute P[48 < T < 50].

(c) Compute P[T > 53.5].

(d) Find the 90th percentile of such temperatures, i.e., find t0 so that P[T t sub 0] = 0.9.

Explanation / Answer

5.6.a)For T1=48, z1=(T1-)/

=(48-50)/4=-0.5

For T2=53.5, z2=(53.5-50)/4=0.88

Thus, P(48<T<53.5)=P(T2<53.5)-P(T1<48)=P(z2<0.88)-P(z1<-0.5)

=0.8106-0.3446=0.466

b)For T2=50, z2=(50-50)/4=0

Thus, P(48<T<50)=P(T2<50)-P(T1<48)=P(z2<0)-P(z1<-0.5)

=0.5-0.3446=0.1556

c)P(T>53.5)=1-P(T<53.5)=1-P(z<0.88)=1-0.8106=0.1894

d)To find 90th percentile find the z score corresponding to 1-0.9=0.1.

The z score is -1.28.

Thus the raw score X=+z=50-1.28*4=44.88

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