Question

Scenario 3 An experimental testing was s carried out in a structures laboratory to determine the critical load under axial compression of struts for a new alternative material. T material. The following data was recorded as the mean critical load for 68 struts Critical 150 157 162 163.5 1661 159 168.3 169.2 171 load (kN) Frequency 7 12 15 Evaluate the mean and standard deviation for this data Considering the struts to be classified into slender and short columns. The mean height for the slender column is given as 160kN with a standard deviation of 7kN. Assuming the load is normally distributed, determine i. The number of struts with reading between 154 and 162kN respectively The number of struts with readings less than 158kN. Assuming 6 struts were separated from the bunch of struts for the test. Determine the probabilities of having: Atleast 1 short column and iv. v. Atleast one short column and one slender column, assuming equal probability of short and slender column in the selection. Sources of information Books SINGH, K. (2011) Engineering Mathematics Through Applications. 2nd Ed. Basingstoke: Palgrave Macmillan STROUD, K. A. and BOOTH, D.J. (2013) Engineering Mathematics. 7th Ed Basingstoke: Palgrave Macmillan. Websites Mathcentre mathcentre.ac.uk Mathtutor mathtutor ac.uk Training/Tutorials)

I need help with Questions iv & v.

Anyone can help me on these please?

Answer 1

P(slender)=P(short)=1/2, there is equal probability of short and slender

(iv) here sample size=n=6 and P(short)=p=1/2=0.5

now we use binomial distribution with parameter n=6 and p=0.5 and for Binomial distribution ,

P(X=r)=ⁿC_rp^r(1-p)^n-r  

P(atleast 1 short column)=P(X>=1)=1-P(X<1)=1-P(X=0)=1-0.0156=0.9844

P(X=0)=0.0156 ( using ms-excel command =binomdist(0,6,0.5,0)

(v) we have 6 column , since 1 short and 1 slender must be selected, remaining would be any type

so required probability=P(one short)*P(one slender)=(1/2)*(1/2)=1/4=0.25