Question

Suppose the diameter at breast height (in.) of a certain type of tree is distributed normally with a mean of 8.8 and a standard deviation of 2.8 as suggested in Aedo-Ortiz, D. M., Olsen, E. D., & Kellogg, L. D. (1997). Simulating a harvester-forwarder softwood thinning: a software evaluation. Forest Products Journal, 47(5), 36.

(a)(2pts) What is the probability that the diameter of a randomly selected tree will be at least 10 inches? Will exceed 10 inches.

(b)(3pts) What is the value of c so that the interval (8.8-c, 8.8+c) contains 98% of all diameter values.

Please provide the answer in Rstudio code. Thank you.

Answer 1

a). For the diameter to be atleast 10 inches, it is one-tailed probability. Hence, lower.tail = FALSE and x = 10

> pnorm(10, mean = 8.8, sd = 2.8, lower.tail = FALSE)
[1] 0.3341176

b). For 98% confidence interval (8.8-c , 8.8+c), = 0.01 (0.02/2) for being two - tailed.

#c = qnorm()*std/sqrt(n) ; std is standard deviation and n is the sample size.

c = qnorm(0.99)*2.8/sqrt(10)
> c
[1] 2.059836

> 8.8-c
[1] 6.740164
> 8.8+c
[1] 10.85984

Therefore, c = 2.059836

and C.I. = (6.740164 , 10.85984)

Note : Since the sample size was not given, I have taken it to be 10 in the formula for calculation of c.