Approximately 99.75% of the students have commute distances between 2.7 miles and 33.3 miles.
We are given;
Mean Distance; x' = 18 miles
Standard deviation; σ = 5.1 miles
Thus;
A) 2.7 = 18 - a(5.1)
where;
a is number of standard deviations away from the mean. Thus;
a = (18 - 2.7)/5.1
a = 3
3 standard deviations from the mean denotes that 99.7% of the data occurs within three standard deviations of the mean within a normal distribution.
B) At 95%, we have the data to be 2 standard deviations from the mean. Thus;
CI = 18 ± 2(5.1)
CI = (18 - 10.2) OR (18 + 10.2)
CI = (7.8, 28.2)
The statements to be completed are;
(a) Approximately ______ of the students have commute distances between 2.7 miles and 33.3 miles.
(b) Approximately 95% of the students have commute distances between _____ miles and ______ miles?
Read more about confidence interval at; https://brainly.com/question/17097944
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