A sample of​ fifth-grade classes was studied in an article. One of the variables collected was the class size in terms of​ student-to-facult

Question

A sample of​ fifth-grade classes was studied in an article. One of the variables collected was the class size in terms of​ student-to-faculty ratio. The​ student-to-faculty ratios of the 70 ​fifth-grade classes sampled have a mean of 17.33 and a standard deviation of 2.36.
Complete parts (a) through (c) below.
a. x-3s
x+s
x-2s
x+2s
x+3s
Type integers or decimals. Do not round.
b. Apply Property 1 of the empirical rule to make pertinent statements about the observations in the sample At least 8 out of 70 fifth-grade classes sampled have student-to-faculty ratios between 15.36 and 17.86 Type integers or decimals. Do not round.
c. Apply Property 2 of the empirical rule to make pertinent statements about the observations in the sample. At least out of 70 fifth-grade classes sampled have student-to-faculty ratios between and Type integers or decimals. Do not round

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4 weeks 2021-12-26T09:53:02+00:00 1 Answer 0 views 0

a)

10.25

19.69

12.61

22.05

24.41

b)

P ( 14.97 < x < 19.69 ) = 0.68

c)

P ( 12.61 < x < 22.05 ) = 0.95

Step-by-step explanation:

Given:

– Sample mean x_bar = 17.33

– Sample mean s.d = 2.36

Find:

x_bar – 3s

x_bar + s

x_bar – 2s

x_bar + 2s

x_bar + 3s

Apply Property 1 of the empirical rule to make pertinent statements about the observations in the sample At least 8 out of 70 fifth-grade classes sampled have student-to-faculty ratios between 15.36 and 17.86

Apply Property 2 of the empirical rule to make pertinent statements about the observations in the sample. At least out of 70 fifth-grade classes sampled have student-to-faculty ratios between and Type integers or decimals.

Solution:

part a)

x_bar – 3s  = 17.33 – 3*2.36 = 10.25

x_bar – 2s  = 17.33 – 2*2.36 = 12.61

x_bar – s  = 17.33 – 2.36 = 14.97

x_bar + 3s  = 17.33 + 3*2.36 = 24.41

x_bar + 2s  = 17.33 + 2*2.36 = 22.05

x_bar + s  = 17.33 + 2.36 = 19.69

part b)

– The first property of empirical rule states that the probability of score between 1 standard deviation from the mean is 68 %. Hence,

P ( x_bar – s.d < x < x_bar + s.d ) = 0.68

P ( 14.97 < x < 19.69 ) = 0.68

part c)

The first property of empirical rule states that the probability of score between 2 standard deviation from the mean is 95 %. Hence,

P ( x_bar – 2*s.d < x < x_bar + 2*s.d ) = 0.95

P ( 12.61 < x < 22.05 ) = 0.95