95% of trees will be within two standard deviations of the mean

99% of trees will be within three standard deviations of the mean

I’ll get you started on question (a). For this one, you need to determine the range that constitutes 95% of the population of tree diameters. 95% (see above) will be within TWO standard deviations of the mean. That means you need to

1) multiply the SD by 2 (i.e., 4.7 x 2)

2) take the figure from above and ADD it to the mean to get the upper limit (i.e., 10.4 + 9.4 = 19.8)

3) take the figure from above and SUBTRACT it from the mean to get the lower limit (i.e., 10.4 – 9.4 = 1.0)

That should give you a good start. Just think it out. Take a look at the normal distribution and that will give you the rest of your answers.

Good luck!

~M~

the standard normal scores for the having the middle 95% of the data is z = ±1.96

a) 10.4 ± 1.96 * 4.7 = (1.188 , 19.612)

b) P(X < 1) = = P(Z < (1 - 10.4) / 4.7) = P(Z < -2) = 0.02275013 c) P(5.7 < X < 10.4) = P(-1 < Z < 0) = P(Z < 0) - P(Z < -1) = 0.50 - 0.1586553 = 0.3413447 d) P(X > 15)

= P(Z > 0.9787)

= 0.1638583

e) P(X < 10.4) = P(Z < 0) = 0.50 look at getting R, it is free software and the most powerful stats software. www.r-project.org