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The mean of the sample mean X that we just calculated is identical to the population mean. We just calculated the standard deviation of the sample mean X, which is the population standard deviation divided by the square root of the sample size: 鈭�10=鈭�20/鈭�2.

(a)

The Sample Mean\(\bar x\)of observations\({x_1},{x_2}, \ldots ,{x_n}\)is calculated as follows:

\(\begin{aligned} \bar x &= \frac{{{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + \ldots + }}{{\rm{x}}_{\rm{n}}}}}{{\rm{n}}}\\&= \frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} \end{aligned}\)

The sample mean can be used to calculate a point estimate of the mean value.

The typical value is estimated to be

\(\begin{aligned} \bar x &= \frac{{{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + \ldots + }}{{\rm{x}}_{\rm{n}}}}}{{\rm{n}}}\\ &= \frac{{\rm{1}}}{{{\rm{16}}}}{\rm{(0}}{\rm{.83 + 0}}{\rm{.88 + \ldots + 1}}{\rm{.83)}}\\ &= 1 {\rm{.3481}}{\rm{.}}\end{aligned}\)

Therefore, the sample mean is \({\rm{1}}{\rm{.3481}}{\rm{.}}\)

(b)

The Sample Median tildex is calculated using \(n\)observations ordered from smallest to greatest, including repeated values. As a result, sample median

\({\rm{\tilde x = }}\)\(\left\{ {\begin{array}{*{20}{l}}{{\rm{ The only middle value if n is odd }}}\\{{\rm{ The average of the two middle values if n is even }}}\end{array}} \right.\)

\(\left\{ {\begin{array}{*{20}{l}}{{{\left( {\frac{{{\rm{n + 1}}}}{{\rm{2}}}} \right)}^{{\rm{th }}}}}\\{{\rm{ average of }}{{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)}^{{\rm{th }}}}{\rm{ and }}{{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)}^{{\rm{th }}}}{\rm{ ordered values , if n is odd }}}\end{array}} \right.\)

The sample median can be used to get the median point estimate.

First and foremost, note that the data has been arranged

\(\begin{array}{l}{\rm{0}}{\rm{.83,0}}{\rm{.88,0}}{\rm{.88,1}}{\rm{.04,1}}{\rm{.09,1}}{\rm{.12,1}}{\rm{.29,1}}{\rm{.31,1}}{\rm{.48,1}}{\rm{.49,1}}{\rm{.59,1}}{\rm{.62,1}}{\rm{.65,1}}{\rm{.71 \backslash \backslash }}\\{\rm{, 1}}{\rm{.76,1}}{\rm{.83 }}{\rm{.}}\end{array}\)

Because this sample has \(n = 16\)observations, which is an even number, the values of interest are

\(\begin{aligned}{\left( {\frac{n}{2}} \right)^{th}}\\ &= {\left( {\frac{{16}}{2}} \right)^{th}}\\ &= {8^{th}}\left( {\frac{n}{2} + 1} \right)\\^{th} &= {\left( {\frac{{16}}{2} + 1} \right)^{th}}\\ &= {9^{th}}\end{aligned}\)

In the ordered sample data, the \({8^{{\rm{th }}}}\)and \({9^{th}}\)values are highlighted in red. As a result, the sample median is the average of the two numbers previously indicated.

\(\begin{array}{c}\tilde x = \frac{{1.31 + 1.48}}{2}\\ = 1.395\end{array}\)

The sample mean might theoretically be employed because of the assumption of normality and the fact that the median is equal to the mean in a normal distribution.

Therefore, the sample median is\(1.395\).

(c)

Percentiles are utilized to solve such situations. The percentile of interest in this case is \({90^{{\rm{th }}}}\)percentile.

As a result, the least \({90^{{\rm{th }}}}\)is separated from the rest by a point estimate of

\(\hat \mu + {z_{1 - \alpha }}\hat \sigma \)

Where\(\hat \mu \)represents the sample mean\({z_{1 - \alpha }}\)represents the \(z - \)of a regular normal distribution, and \(\hat \sigma \)represents the sample standard deviation.

The sample mean has been calculated, and the z-score \(\alpha = 0.1 = 10\% \) can be found in the book's appendix (standard normal distribution).

\(\begin{array}{c}{\rm{\hat \mu = \bar x}}\\{\rm{ = 1}}{\rm{.3481}}{{\rm{z}}_{{\rm{1 - \alpha }}}}\\{\rm{ = }}{{\rm{z}}_{{\rm{1 - 0}}{\rm{.1}}}}\\{\rm{ = 1}}{\rm{.28}}\end{array}\)

The sample standard deviation is the only thing that has to be calculated.

\({s^2}\)is the Sample Variance.

\({{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{n - 1}}}}{\rm{ \times }}{{\rm{S}}_{{\rm{xx}}}}\)

Where,

\(\begin{array}{c}{{\rm{S}}_{{\rm{xx}}}}{\rm{ = }}\sum {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - \bar x}}} \right)}^{\rm{2}}}} \\{\rm{ = }}\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}\end{array}\)

\(s\)is the Sample Standard Deviation.

\(\begin{array}{c}{\rm{s = }}\sqrt {{{\rm{s}}^{\rm{2}}}} \\{\rm{ = }}\sqrt {\frac{{\rm{1}}}{{{\rm{n - 1}}}}{\rm{ \times }}{{\rm{S}}_{{\rm{xx}}}}} \end{array}\)

The sample variance is the point estimate of the variance, while the sample standard deviation is the point estimate of the standard deviation.

\(x_i^2\)are the squared data points.

\(0.6889,0.7744,0.7744,1.0816,1.1881,1.2544,1.6641,1.7161,2.1904,2.2201,2.5281,2.6244,2.7225,2.9241,3.0976,3.3489{\rm{, }}\)

Thus, the \({S_{xx}}\)

\(\begin{array}{c}{{\rm{S}}_{{\rm{xx}}}}{\rm{ = }}\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}\\{\rm{ = 0}}{\rm{.6889 + 0}}{\rm{.7744 + \ldots + 3}}{\rm{.3489 - }}\frac{{\rm{1}}}{{{\rm{16}}}}{\rm{ \times (0}}{\rm{.83 + 0}}{\rm{.88 \ldots + 1}}{\rm{.83}}{{\rm{)}}^{\rm{2}}}\\{\rm{ = 30}}{\rm{.7981 - }}\frac{{\rm{1}}}{{{\rm{16}}}}{\rm{ \times 21}}{\rm{.57}}\\^{\rm{2}}{\rm{ = 1}}{\rm{.719}}\end{array}\)

And the sample variance is

\(\begin{array}{c}{{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{16 - 1}}}}{\rm{ \times 1}}{\rm{.719}}\\{\rm{ = 0}}{\rm{.1146}}\end{array}\)

And the sample standard deviation is

\(\begin{array}{c}{\rm{s = }}\sqrt {{{\rm{s}}^{\rm{2}}}} \\{\rm{ = }}\sqrt {{\rm{0}}{\rm{.1146}}} \\{\rm{ = 0}}{\rm{.3385}}\end{array}\)

Finally, the \({90^{{\rm{th }}}}\)percentile is

\(\begin{array}{c}{\rm{\hat \mu + }}{{\rm{z}}_{{\rm{1 - \alpha }}}}{\rm{\hat \sigma = \bar x + }}{{\rm{z}}_{{\rm{0}}{\rm{.9}}}}{\rm{ \times s}}\\{\rm{ = 1}}{\rm{.3481 + 1}}{\rm{.28 \times 0}}{\rm{.3385}}\\{\rm{ = 1}}{\rm{.7814}}{\rm{.}}\end{array}\)

Such that,the \({90^{{\rm{th }}}}\)percentile is \({\rm{1}}{\rm{.7814}}{\rm{.}}\)

Now, calculation of Sample mean

Percentiles are utilized to solve such situations. The percentile of interest in this case is \({90^{{\rm{th }}}}\)percentile. As a result, the least \({90^{{\rm{th }}}}\)is separated from the rest by a point estimate of

\({\rm{\hat \mu + }}{{\rm{z}}_{{\rm{1 - \alpha }}}}{\rm{\hat \sigma }}\)

Where hat represents the sample mean, \({z_{1 - \alpha }}\)represents the \(z\)of a regular normal distribution, and \(\hat \sigma \)represents the sample standard deviation.

The sample mean has been calculated, and the \(z\)for \({\rm{\alpha = 0}}{\rm{.1 = 10\% }}\)can be found in the book's appendix (standard normal distribution).

\(\begin{array}{c}{\rm{\hat \mu = \bar x}}\\{\rm{ = 1}}{\rm{.3481}}{{\rm{z}}_{{\rm{1 - \alpha }}}}\\{\rm{ = }}{{\rm{z}}_{{\rm{1 - 0}}{\rm{.1}}}}\\{\rm{ = 1}}{\rm{.28}}\end{array}\)

The sample standard deviation is the only thing that has to be calculated.

Estimation of Sample Variance:

The sample variance\({s^2}\)is

\({{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{n - 1}}}}{\rm{ \times }}{{\rm{S}}_{{\rm{xx}}}}\)

Where,

\(\begin{array}{c}{\rm{s = }}\sqrt {{{\rm{s}}^{\rm{2}}}} \\{\rm{ = }}\sqrt {\frac{{\rm{1}}}{{{\rm{n - 1}}}}{\rm{ \times }}{{\rm{S}}_{{\rm{xx}}}}} \end{array}\)

The sample variance is the point estimate of the variance, while the sample standard deviation is the point estimate of the standard deviation. The squared data points, \(x_i^2\), are\(0.6889,0.7744,0.7744,1.0816,1.1881,1.2544,1.6641,1.7161,2.1904,2.2201,2.5281,2.6244,2.7225,2.9241,3.0976,3.3489{\rm{, }}\)

Thus, the \({\rm{S\_\{ x x\} }}\)is

\(\begin{array}{c}{{\rm{S}}_{{\rm{xx}}}}{\rm{ = }}\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}\frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times }}{\left( {\sum {{{\rm{x}}_{\rm{i}}}} } \right)^{\rm{2}}}\\{\rm{ = 0}}{\rm{.6889 + 0}}{\rm{.7744 + \ldots + 3}}{\rm{.3489 - }}\frac{{\rm{1}}}{{{\rm{16}}}}{\rm{ \times (0}}{\rm{.83 + 0}}{\rm{.88 \ldots + 1}}{\rm{.8}}{{\rm{3}}^{\rm{2}}}\\{\rm{ = 30}}{\rm{.7981 - }}\frac{{\rm{1}}}{{{\rm{16}}}}{\rm{ \times 21}}{\rm{.5}}{{\rm{7}}^{\rm{2}}}\\{\rm{ = 1}}{\rm{.719}}\end{array}\)

And the sample variance is:

\(\begin{array}{c}{{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{16 - 1}}}}{\rm{ \times 1}}{\rm{.719}}\\{\rm{ = 0}}{\rm{.1146}}\end{array}\)

And the standard deviation of the sample is

\(\begin{array}{c}s = \sqrt {{s^2}} \\ = \sqrt {0.1146} \\ = 0.3385\end{array}\)

Finally, the \({90^{{\rm{th }}}}\)percentile is

\(\begin{array}{c}\hat \mu + {z_{1 - \alpha }}\hat \sigma = \bar x + {z_{0.9}} \cdot s\\ = 1.3481 + 1.28 \cdot 0.3385\\ = 1.7814\end{array}\)

Therefore, the , the \({90^{{\rm{th }}}}\)percentile is \(1.7814\)

\(({\rm{d}}):\)

The estimate of the probability of interest is due to the use of the mean and standard deviation estimations.

\(\begin{array}{c}P(X < 1.5) = P\left( {\frac{{X - \bar x}}{s} < \frac{{1.5 - 1.3481}}{{0.3385}}} \right)\\ = P(Z < 0.45)\\\mathop = \limits^{(1)} 0.6737\end{array}\)

(1): from the appendix's normal probability table. A software can also be used to calculate the probability.

Therefore, the mean and standard deviation estimations, \({\rm{P(X < 1}}{\rm{.5) = 0}}{\rm{.6737 ;}}\)

(e):

Due to normality, the estimator \(X's\)predicted standard error is \(X's\)

\(\begin{array}{c}{{\rm{S}}_{{\rm{\bar x}}}}{\rm{ = }}\frac{{{\rm{\hat \sigma }}}}{{\sqrt {\rm{n}} }}\\{\rm{ = }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.3385}}}}{{\sqrt {{\rm{16}}} }}\\{\rm{ = 0}}{\rm{.0846}}\end{array}\)

Hence, the Estimated standard error is\({\rm{0}}{\rm{.0846}}\).