## 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

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

## Answers ( )

Answer: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 ) = 0.68part 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