Z = (x – μ) / σ
# finding value for 'x': x <- Z * sd + mu
mu <- 0
sd <- 1
#a
Z <- -1.13
x <- Z * sd + mu
# prob of x > -1.13
1 - pnorm(x, mean = 0, sd = 1)
## [1] 0.8707619
#b
Z <- 0.18
x <- Z * sd + mu
# prob of x < 0.18
pnorm(x, mean = 0, sd = 1)
## [1] 0.5714237
#c
Z <- 8
x <- Z * sd + mu
# prob of x > 8
1 - pnorm(x, mean = 0, sd = 1)
## [1] 6.661338e-16
#d
Z <- 0.5
x <- Z * sd + mu
# prob of |x| < 0.5 = -x < 0.5 < x
x1 <- pnorm(-x, mean = 0, sd = 1)
x2 <- pnorm(x, mean = 0, sd = 1)
x2 - x1
## [1] 0.3829249
# shadeDist function plots a probability density function, shades the area under the curve, and computes the probability.
mean <- 0
SD <- 1
x <- seq(-4, 4, length = 10000)
y <- dnorm(x, mean, SD)
par(mfrow=c(2,2))
shadeDist(-1.13, lower.tail = FALSE, col = c("blue", "light blue"))
shadeDist(0.18, col = c("blue", "light blue"))
shadeDist(8, lower.tail = FALSE, col = c("blue", "light blue"))
shadeDist(c(-0.5,0.5), lower.tail = FALSE, col = c("blue", "light blue"))
the percent of standard normal distribution found in a: 87.08% the percent of standard normal distribution found in b: 42.86% the percent of standard normal distribution found in c: 0% the percent of standard normal distribution found in d: 38.29%
# *Z = (x – μ) / σ*
m.mean <- 4313
m.sd <- 583
leo <- 4948
w.mean <- 5261
w.sd <- 807
mary <- 5513
#(b)
leo.z <- (leo - m.mean) / m.sd
mary.z <- (mary - w.mean) / w.sd
#(c)
leo.rank <- 1 - pnorm(leo, m.mean, m.sd)
mary.rank <- 1 - pnorm(mary, w.mean, w.sd)
#(d)
leo.top <- 1 - pnorm(leo, m.mean, m.sd)
mary.top <- 1 - pnorm(mary, w.mean, w.sd)
colnames(fheights) <- "heights"
heights <- fheights$heights
h.mean <- 61.52 # mean(fheights$heighs)
h.sd <- 4.58 # sd(fheights$heighs)
# (a)
# Within 1 standard deviation of the mean.
length(which(heights > (h.mean - h.sd) & heights < (h.mean + h.sd)))/length(heights)
## [1] 0.6666667
# Within 2 standard deviation of the mean.
length(which(heights > (h.mean - 2 * h.sd) & heights < (h.mean + 2 * h.sd)))/length(heights)
## [1] 0.9583333
# Within 3 standard deviation of the mean.
length(which(heights > (h.mean - 3 * h.sd) & heights < (h.mean + 3 * h.sd)))/length(heights)
## [1] 1
# (b)
par(mfrow = c(1,2))
hist(fheights$heights)
qqnorm(fheights$heights)
qqline(fheights$heights)
source("./qqnormsim.R")
qqnormsim(fheights$heights)
#success: finding defective part
a <- (1 - 0.02)^9 * 0.02
0.016675 chance for 1 success at 2% defective rate after 9 failures. (b)
b <- 1 - 0.98^100
b
## [1] 0.8673804
0.8673804 of chance of 100 succeses in a row. (c)
#E(x) = 1/p
c.mean <- 1/.02
c.sd <- ((1 - .02)/.02^2)^(1/2)
On average, 50 transistors to expect before the first one to be produced with a defect, with a standard deviation of 49.4974747 (d)
d.mean <- 1/.05
d.sd <- ((1 - .05)/.05^2)^(1/2)
On average, I would expect 20 transistor to be produced with the first defect, with a standard deviation of 19.4935887 (e) Increasing the probability of an event decreases the mean and standard deviation of the wait time until the event. Lower chance of success means more trials until success happens. If the event doesn’t happen for 100 days, the probability of the event happening on the 101st day is still at 2% and 5%.
n <- 3
k <- 2
p <- 0.51
double_boy <- choose(n, k) * (1 - p)^(n - k) * (p)^k
double_boy
## [1] 0.382347
(b)
#(B,B,G), (B,G,B), (G,B,B)
b <- 0.51
g <- 0.49
(b*b*g) + (b*g*b) + (g*b*b)
## [1] 0.382347
(c) A and B are just different ways of writing the same mathematical operations. The use of the formula is quicker way to get the result to save time. In this case, the second methold will require us to create combination of 56 different possibilities.
#Negative binomial dist.
p <- 0.15
n <- 10
k <- 3
choose(n - 1, k - 1) * (1 - p)^(n - k) * p^k
## [1] 0.03895012
(b) 15%. All serves are independent, thus the probability of the next serve is still at 15% of chance. (c) The given information in B, “two sucessful serves in nine attempts” is not affecting the probability of sucess on the next serve as they are independent disjoint events. The calculation of A,B are in different sense that it is not reaonsable to compare the two.