Respuesta :
Answer: Fourth option is correct.
Step-by-step explanation:
Since we have given that
Mean of sample = 65.3
Standard deviation = 5.2
We need to find the 90% confidence interval for the mean.
So, z = 1.64
And the interval would be
[tex]\bar{x}\pm z\times \dfrac{\sigma}{\sqrt{n}}\\\\=65.3\pm 1.64\times \dfrac{5.2}{\sqrt{36}}\\\\=65.3\pm 1.42\\\\=(65.3-1.42,65.3+1.42)\\\\=(63.88,66.72)\\\\=(63.9,66.7)[/tex]
Hence, Fourth option is correct.
Using the t-distribution, it is found that the 90% confidence interval for the mean height of all female students in their school is (63.8, 66.8).
We are given the standard deviation for the sample, thus, the t-distribution is used to solve this question.
For the sample, we have that:
- 36 students, thus [tex]n = 36[/tex].
- Mean of 65.3, thus [tex]\overline{x} = 65.3[/tex].
- Standard deviation of 5.2, thus [tex]s = 5.2[/tex].
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
The first step is finding the number of degrees of freedom, which is 1 less than the sample size, thus df = 36.
Then, using a calculator or the t-table, we find the critical value for a 90% confidence interval with 36 df, which is t = 1.6896.
Then:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 65.3 - 1.6896\frac{5.2}{\sqrt{36}} = 63.8[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 65.3 + 1.6896\frac{5.2}{\sqrt{36}} = 66.8[/tex]
The 90% confidence interval for the mean height of all female students in their school is (63.8, 66.8).
A similar problem is given at https://brainly.com/question/15872108
