adult tickets cost $10 and children's tickets to a show cost $5. if the number of children's tickets was twice the number of adu
1 answer:
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Answer:
1). Z score = 1.464733
Mean = 38.87917
Standard deviation = 3.609405
2). Quartiles:
Q1 = 37.025
Q2 = 38.450
Q3 = 40.800
3). IQR = 3.775
4).
CI (lower fence) = 37.4351
CI (upper fence) = 40.32323
5). There is outlier in the data set. Please see attached box plot for evidence.
Step-by-step explanation:
1). By Z score, we mean:
Z = ,
where:
x-bar ==> mean(g) = 38.87917
= 37.80
standard deviation = 3.609405
sample size (n) = 24.
2). By quantile, we mean:
Q = L + (i*(n/4) - Cf)*c
Where L is the lower class limit of the quartile class
Cf is the cumulative frequency before the quartile class
c is the class size.
3) . IQR = Q3 - Q1
4). CI = ,
Where ==> 1.96
In order to replicate and obtain the result, please use the R code below:
g = c(31.5, 36.0, 37.8, 38.5, 40.1, 42.2,34.2, 36.2, 38.1, 38.7, 40.6, 42.5,34.7, 37.3, 38.2, 39.5,
41.4, 43.4,35.6, 37.6, 38.4, 39.6, 41.7, 49.3)
boxplot(g)
Z = (mean(g) - 37.8)/(sd(g)/sqrt(length(g)))
mean(g)
quantile(g)
IQR(g)
CIl = mean(g) - 1.96*(sd(g)/sqrt(length(g)))
CIU = mean(g) + 1.96*(sd(g)/sqrt(length(g)))
