Question

reyesburg corporation is contemplating using a new, more expensive glue in the construction of its laminated veneer lumber. of importance to the company is the carrying load of the lumber. the company has tested 27 beams using the new glue, recording for each beam the pressure (in pounds per square foot) at which the beam breaks. the data collected are presented in the following frequency distribution.

pressure (in pounds per square foot)	frequency
880 to 899	7
900 to 919	10
920 to 939	3
940 to 959	2

based on the frequency distribution, using the midpoint of each data class, estimate the mean carrying load of the beams tested. for your intermediate computations, use four or more decimal places, and round your answer to one decimal place.

Explanation:

Step1: Find midpoints of each class

For class 860 - 879: Midpoint \( x_1=\frac{860 + 879}{2}=869.5 \)
For class 880 - 899: Midpoint \( x_2=\frac{880 + 899}{2}=889.5 \)
For class 900 - 919: Midpoint \( x_3=\frac{900 + 919}{2}=909.5 \)
For class 920 - 939: Midpoint \( x_4=\frac{920 + 939}{2}=929.5 \)
For class 940 - 959: Midpoint \( x_5=\frac{940 + 959}{2}=949.5 \)

Step2: Calculate \( f_i\times x_i \) for each class

• For 860 - 879: \( f_1 = 5 \), \( f_1x_1=5\times869.5 = 4347.5 \)
• For 880 - 899: \( f_2 = 7 \), \( f_2x_2=7\times889.5 = 6226.5 \)
• For 900 - 919: \( f_3 = 10 \), \( f_3x_3=10\times909.5 = 9095 \)
• For 920 - 939: \( f_4 = 3 \), \( f_4x_4=3\times929.5 = 2788.5 \)
• For 940 - 959: \( f_5 = 2 \), \( f_5x_5=2\times949.5 = 1899 \)

Step3: Find total of \( f_i \) and \( f_i\times x_i \)

Total \( f=\sum f_i=5 + 7+10 + 3+2=27 \)
Total \( \sum f_ix_i=4347.5+6226.5 + 9095+2788.5+1899=24356.5 \)

Step4: Calculate the mean

Mean \( \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{24356.5}{27}\approx902.1 \) (rounded to one decimal place)

Answer:

\( 902.1 \)