In a facility that manufactures electrical resistors, a sample of 35 1-k 2 resis

Question

In a facility that manufactures electrical resistors, a sample of 35 1-k 2 resistors are randomly pulled from the production line, and their resistances are measured and recorded, as shown in the below. The desired resistance tolerance for the resistor is10%, meaning that the acceptable range of resistance is 900 to 1 100 a. Create a histogram of the data. b. Calculate the mean, median and mode c. Calculate the standard deviation (assume the data represents a sample) d. Assume a normal distribution. How many standard deviations on either side of the mean does the ± 10% tolerance represent? Assuming a normal distribution, what percentage of resistors manufactured by that process will be out of specification? e. Measured Resistance (2) 1001.6 931.5 1036.4990.7979.41021.5 9068 981.7 919.1998.6 950.0 940.7 912.0 836.9 877.5 962.7 997.6 1094.6 921.6 939.4 898.6 980.0 859.8 944.1 984.31176.8 987.6 1023.4 941.8 1067.51071.3 958.8 1115.9 1055.4 1027.4

Explanation / Answer

ans)

x=c(1001.6,931.5,1036.4,990.7,979.4,1021.5,906.8,981.7,919.1,998.6,950.0,940.7,
912.0,836.9,877.5,962.7,997.6,1094.6,921.6,939.4,898.6,980.0,859.8,944.1,
984.3,1176.8,987.6,1023.4,941.8,1067.5,1071.3,958.8,1115.9,1055.4,1027.4)

#a)hist
h=hist(x)

#b)mean,median,mode
m1=mean(x) : mean is 979.8
m2=median(x) : meadian is 980
library(modeest)
m3=mlv(x, method = "mfv") : mode is 979.8

#c)standard deviation
s1=sd(x) : standard deviation is 73.05499

#d)how many sd

z1=(900-m1)/s1
z2=(1100-m1)/s1

Here we need to calculate the Z-scores for two tolerances ie for 900 and 1100. The Z-score represents how many standard deviation is above or below the mean. The z-score for 900 is -1.092328 which is negative means that the point is below the mean. The Z-score for 1100 is 1.645336 which is positive means the point is above the mean. The above Z-scores represents the number of standard deviation away from the mean.

#e) percentage
z=(x-m1)/s1
x1=length(which(z<z1 | z>z2))
p=(x1/35)*100

Here we need to calculate the proportion of Z-scores that lies outside the Z-scores of -10% and -10% tolerance limits.The given proportion multiplied by 100 will give the percentage of resistors manufactured by the process that will be out of specification. Here the value is 17.14286%

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