The diameters of ball bearings produced at a factory are approximately normally
ID: 3132558 • Letter: T
Question
The diameters of ball bearings produced at a factory are approximately normally distributed. Suppose the mean diameter is 1.002 centimeters cm and the standard deviation is 0.006 cm. The product specifications require that the diameter of each ball bearing be between 0.980 and 1.020 cm. (hint: use table 5 in back of your text and either function NORM.DIST or NORM.INV as appropriate). a) What proportion of ball bearings can be expected to have a diameter under 1.020 cm? b) What proportion of ball bearings can be expected to have a diameter over 1.020 cm? c) What proportion of ball bearings can be expected to have a diameter between 0.980 and 1.020 cm? That is what proportion of ball bearings can be expected to meet the specifications? d) What is the probability that the diameter of a randomly selected ball bearing will be over 1.000 cm? e) What is the probability that the diameter of a randomly selected ball bearing will be under 0.995 cm? f) What is the percentile rank of a ball bearing that has a diameter of 0.991? g) What is the percentile rank of a ball bearing that has a diameter of 1.011 cm? h) Determine the 10th percentile of diameters of ball bearings i) Determine the diameters of ball bearings that make up the middle 99% of all diameters.
Explanation / Answer
Normal Distribution
The diameters of ball bearings produced at a factory are approximately normally distributed. Suppose the mean diameter is 1.002 centimetres cm and the standard deviation is 0.006 cm. The product specifications require that the diameter of each ball bearing be between 0.980 and 1.020 cm. (hint: use table 5 in back of your text and either function NORM.DIST or NORM.INV as appropriate).
We are given that the diameters of ball bearings are normally distributed.
Mean = 1.002
Standard deviation = 0.006
Specification limits = 0.980 to 1.020
Solution:
Here, we have to find P(X<1.020)
Z = (1.020 – 1.002) / 0.006
Z = 3
P(X<1.020) = P(Z<3) = 0.99865
Required probability = 0.99865
Solution:
Here, we have to find P(X>1.020)
Z = (1.020 – 1.002) / 0.006
Z = 3
P(X>1.020) = 1 - P(X<1.020) =1 - P (Z<3) =1 - 0.99865 = 0.00135
Required probability = 0.00135
Solution:
Here, we have to find P(0.980<X<1.020)
P(0.980<X<1.020) = P(X<1.020) – P(X<0.998)
P(X<1.020)
Z = (1.020 – 1.002) / 0.006
Z = 3
P(X<1.020) = P(Z<3) = 0.99865
P(X<0.980)
Z = (0.980 – 1.002) / 0.006
Z = -3.6667
P(X<0.980) = P(Z<-3.6667) = 0.000123
P(0.980<X<1.020) = P(X<1.020) – P(X<0.998) = P(Z<3) – P(Z<-3.6667)
P(0.980<X<1.020) = 0.99865 – 0.000123 = 0.998527
Required probability = 0.998527
Solution:
Here, we have to calculate P(X>1.000)
Z = (1.000 – 1.002) / 0.006 = -0.3333
P(X>1.000) = 1 – P(X<1.000) = 1 – P(Z<-0.3333) = 1 – 0.3694414 = 0.6305587
Required probability = 0.6305587
Solution:
Here, we have to calculate P(X<0.995)
Z = (0.995 – 1.002) / 0.006
Z = -1.6667
P(X<0.995) = P(Z<-1.6667) = 0.1216725
Required probability = 0.1216725
Solution:
Here, we have to find percentile rand for diameter 0.991
Z = (0.991 – 1.002) / 0.006 = -1.8333
P(Z<-1.8333) = 0.0333765
Percentile rank = 3.33% or approximately 4th percentile
Solution:
Z = (1.011 – 1.002) / 0.006 = 1.5
P(Z<1.5) = 0.9332
Percentile rank = 93.32% or approximately 94th percentile
Solution:
Z value for 10th percentile or 0.10 = -1.281552
X = Mean + Z*SD
X = 1.002 + (-1.281552)*0.006
X = 1.002 – 1.281552*0.006
X = 0.9943107
Required answer = 0.9943
Solution:
Here, we have to find the diameters for the lower 1% and upper 99% of all diameters.
Z value for lower 1% = -2.326348
Z value for upper 99% = 2.326348
Lower diameter = X = Mean + Z*SD = 1.002 – 2.326348*0.006
Lower diameter = 0.9880
Upper diameter = X = Mean + Z*SD = 1.002 + 2.326348*0.006
Upper diameter = 1.0159581
Required answer = all diameters between 0.9880 and 1.0160