Test of the stiffness of a number of aluminum alloy channels

Test of the stiffness of a number of aluminum alloy channels gave the following frequency distribution Stillness was measured in \"effective El in lb/in^2 \" Use the long method to compute X and s. If a normal distribution had these values for mu and sigma. what percentage of the distribution would fall below 2.150?

Solution

Let x be stiffness cell mid point and f be the frequency

Mean= 2400; SD =38.16

The percentage of values below 2150 is P(X<2150) = P(Z<(2150-2400/38.16))=P(Z<-6.54)= 0.000

The precentage of values below 2150 is approximately 0%.

Mid points-x	frequency-f	x*f	(x-mean)^2
2640	1	2,640	57,792
2600	2	5,200	40,160
2560	7	17,920	25,728
2520	11	27,720	14,496
2480	25	62,000	6,464
2440	33	80,520	1,632
2400	41	98,400	0
2360	35	82,600	1,568
2320	22	51,040	6,336
2280	14	31,920	14,304
2240	5	11,200	25,472
2200	3	6,600	39,840
2160	1	2,160	57,408
Sum	200	479,920	291,202
Mean=	Sum x*f/Sum-f	2,400
Variance=	sum(x-mean)^2/Sum-f	1,456
Standard deviation =	sqrt(Variance)	38.15770433