Question

Suppose the value of Young's modulus (GPa) was determined for cast plates consisting of certain intermetallic substrates, resulting in the following sample observations:

116.2

115.9

114.8

115.3

115.7

(a) Calculate x. (Enter your answer to two decimal places.)
GPa

Calculate the deviations from the mean. (Enter your answers to two decimal places.)

x	116.2	115.9	114.8	115.3	115.7
deviation

(b) Use the deviations calculated in part (a) to obtain the sample variance and the sample standard deviation. (Round your answers to three decimal places.)

s2	=	GPa2
s	=	GPa

(c) Subtract 50 from each observation to obtain a sample of transformed values. Now calculate the sample mean and sample variance of these transformed values. (Round your answer to three decimal places.)
x_t = GPa
s_t² = GPa²

Compare the transformed x and s² to the original data.

The transformed mean and variance are both the same as the original.The transformed mean is less than the original and the transformed variance is less than original variance.    The transformed mean is less than the original but the transformed variance is the same original variance.The transformed mean is the same as the original but the transformed variance is less than the original variance.The transformed mean is less than the original and the transformed variance is greater than original variance.

Answer 1

S.No		X	|X-x̄|	(X-x̄)2
1		116.200	0.620	0.38440
2		115.900	0.320	0.10240
3		114.800	-0.780	0.60840
4		115.300	-0.280	0.07840
5		115.700	0.120	0.01440
Σx		577.9	Σ(X-x̄)2=	1.1880
x̄=Σx/n		115.5800	s2=Σ(x-x̄)2/(n-1)=	0.297
			s=√s2    =	0.545

from above:

a)

sample mean xbar =115.58

x	116.2	115.9	114.8	115.3	115.7
deviation	0.62	0.32	-0.78	-0.28	0.12

b)

s² =0.297

s =0.545

c)

xt =65.58

s² =0.297

The transformed mean is less than the original but the transformed variance is the same original variance