Question

Suppose a researcher believes that the average height of female students in a large local high school is 140 cm. The researcher wants to construct an interval that contains the true average height of all female students in the local high school with a certain prespecified probability. The researcher selects 36 female students at random from the high school. The distribution of heights is known to follow a normal distribution.

Suppose a researcher believes that the average height of female students in a large local high school is 140 cm. The researcher wants to construct an interval that contains the true average height of all female students in the local high school with a certain prespecified probability. The researcher selects 36 female students at random from the high school. The distribution of heights is known to follow a normal distribution. The given table shows the data she collected (in cm). O 141 128 140 134 147 135 139 142 144 138 138 143 146 146 140 140 140 141 140 139 150 141 138 145 138 136 140 142 140 137 141 142 145 144 136 146 The average height of the female students in the given sample is cm. The sample standard deviation is cm. (Round your answers to two decimal places.)

Answer 1

Solution:-

X	(X – Mean)^2
139	2.59
146	29.05
141	0.15
140	0.37
134	43.69
138	6.81
142	1.93
146	29.05
128	159.01
139	2.59
147	40.83
136	21.25
144	11.49
140	0.37
140	0.37
150	88.17
135	31.47
140	0.37
138	6.81
140	0.37
141	0.15
141	0.15
142	1.93
142	1.93
138	6.81
140	0.37
144	11.49
138	6.81
145	19.27
140	0.37
143	5.71
141	0.15
136	21.25
145	19.27
146	29.05
137	13.03
5062	614.48

(A). Average height = Sum of X / No of Female Student

                               = 5062 / 36

                               = 140.61

(B). SD = √ ((sum ((X-mean)^2))/(n-1))

             = √ ((614.48 / (36-1)))

             = √ 17.55657

             = 4.19

So, the sample standard deviation is 4.19