Answer:
The answer is explained below
Step-by-step explanation:
stiffness frequency
2480 23
2440 35
2400 40
2360 33
2320 21
a) The mean for the population is calculated using the formula:
[tex]mean (\mu)=\frac{\Sigma f_ix_i}{\Sigma f_i} \\=\frac{x_1f_1+x_2f_2+.\ .\ .+x_nf_n}{x_1+x_2+.\ .\ . +x_n} \\=\frac{(2480*23)+(2440*35)+(2400*40)+(2360*33)+(2320*21)}{23+35+40+33+21} =\frac{365040}{152}=2401.6[/tex]
b) The standard deviation is given by:
[tex]\sigma=\sqrt{ \frac{\Sigma f_i(x_i-\mu)^2}{\Sigma f_1} } \\=\sqrt{ \frac{f_1(x_1-\mu)^2+f_2(x_2-\mu)^2+.\ .\ .+f_n(x_n-\mu)^2}{f_1+f_2+.\ .\ .+f_n} } \\=\sqrt{ \frac{23(2480-2401.6)^2+35(2440-2401.6)^2+40(2400-2401.6)^2+33(2360-2401.6)^2+21(2320-2401.6)^2}{23+35+40+33+21 }}\\=\sqrt{\frac{390021.12}{152} }= 50.7[/tex]
c) We have to find the z score for x = 2350. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma}=\frac{2350-2401.6}{50.7}=-1.02[/tex]
From the z table:
The probability that stiffness would be less than 2350 for any given channel section = P(x < 2350) = P(z < -1.02) = 0.1539 = 15.39%
