91Ó°ÊÓ

The brain weights of Swedish men follow a normal distribution with a mean of $1.40$ and a standard deviation of $0.11$

Formula used: Standard Deviation$={\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

Let $x$ represent Swedish men's brain weights. Then, as follows, $x$ follows the Normal distribution: $x~N(1.40,0.11)$

With a mean of $\mathrm{μ}$and a standard deviation of $\mathrm{σ}$the population distance is normally distributed. The sample mean sampling distribution for sample size $3$. The sampling distribution $\stackrel{¯}{x}$will be normal, with a mean of ${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$and a standard deviation of ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$, where $n$is the sample size.

is then determined as follows:

Here the sample size $n=3$

Mean of$\stackrel{¯}{x}$is${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}=1.40$

Standard deviation of $\stackrel{¯}{x}$is,

${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

$$\begin{gathered} =\frac{0.11}{\sqrt{3}} \\ =0.0635 \end{gathered}$$

As a result, the sampling distribution becomes $\stackrel{¯}{x}~N(1.40,0.0635)$

As a result, the brain weights in the sample exhibit a normal distribution, with a mean of $1.40\mathrm{kg}$and a standard deviation of $0.064\mathrm{kg}$

With a mean of $\mathrm{μ}$and a standard deviation of $\mathrm{σ}$, the population distance is normally distributed. The sample mean sampling distribution for sample size $12$is then determined as follows:

The sampling distribution $\stackrel{¯}{x}$will be normal as well, with a mean of ${\mathrm{μ}}_{\mathrm{λ}}=\mathrm{μ}$and a standard deviation of ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$, where $n$is the sample size.

Here the sample size $n=12$

Mean of $\stackrel{¯}{x}$is, ${\mathrm{μ}}_{\stackrel{˙}{x}}=\mathrm{μ}=1.40$

Standard deviation of $\stackrel{¯}{x}$is,

${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

$$\begin{gathered} =\frac{0.11}{\sqrt{12}} \\ =0.0317 \end{gathered}$$

As a result, the sampling distribution can be calculated as $\stackrel{¯}{x}~N(1.40,0.0317)$

As a result, the mean brain weights in the sample have a normal distribution, with a mean of $1.40\mathrm{kg}$ and a standard deviation of $0.032\mathrm{kg}$

Using Minitab, create the graphs as follows:

• Open Minitab, go to Graph, and select Probability distribution plot from the drop-down menu.
• To continue, select View Single Plot and click Ok.
• Select the Normal distribution and fill in the Mean and Standard Deviations.
• To create the graph, click Ok.

Let $\mathrm{μ}$be the average Swedish man's brain weight. The sample size is $n=3$based on the information provided. The sample mean brain weights of Swedish males are thus designated by $\stackrel{¯}{x}$, and they are roughly distributed with ${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$and ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}=\frac{0.1}{\sqrt{3}}=0.064$

The percentage of all three Swedish males whose mean brain weights are within $0.1\mathrm{kg}$of the population mean brain weight of $1.40\mathrm{kg}$is then about equivalent to the normal curve with parameters $1.40and0.064$and sits between $\mathrm{μ}-0.1$and $\mathrm{μ}+0.1$The associated $z-$scores are then,

$$\begin{gathered} z=\frac{(\mathrm{μ}-0.1)-\mathrm{μ}}{0.064}z=\frac{(\mathrm{μ}+0.1)-\mathrm{μ}}{0.064} \\ z=\frac{0.1}{0.064}z=\frac{0.1}{0.064} \\ z=-1.56z=1.56 \end{gathered}$$

The area under the standard normal curve between $-1.56$and $1.56$is, according to Table II.

$\mathrm{Φ}(1.56)-\mathrm{Φ}(-1.56)=0.9406-0.0594=0.8812$

As a result, the mean brain weights of the three Swedish guys were within $0.1\mathrm{kg}$ of the population mean brain weight of $1.40\mathrm{kg}$ in $88.12\%$of the samples.

Let $\mathrm{μ}$be the average Swedish man's brain weight. The sample size is $n=12$based on the information provided. The sample mean brain weights of Swedish males are thus designated by $\stackrel{¯}{x}$, and they are roughly distributed with ${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$and ${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}=\frac{0.1}{\sqrt{12}}=0.032$

The fraction of $12$ Swedish guys with mean brain weights within $0.1\mathrm{kg}$ of the population average of $1.40\mathrm{kg}$ is then about equivalent to the normal curve with parameters $1.40$and $0.032$and sits between $\mathrm{μ}-0.1$and $\mathrm{μ}+0.1$The associated $z-$scores are then,

$$\begin{gathered} z=\frac{\left(\mathrm{μ}-0.1\right)-\mathrm{μ}}{0.032}z=\frac{\left(\mathrm{μ}+0.1\right)-\mathrm{μ}}{0.032} \\ z=\frac{-0.1}{0.032}z=\frac{0.1}{0.032} \\ =-3.13=3.13 \end{gathered}$$

The area under the standard normal curve between $-1.56$and $1.56$is, according to Table II.

$$\begin{gathered} \mathrm{Φ}(3.13)-\mathrm{Φ}(-3.13)=0.9991-0.0009 \\ =0.9982 \end{gathered}$$

As a result, the mean brain weights of the three Swedish guys are within $0.1\mathrm{kg}$of the population mean brain weight of $1.40\mathrm{kg}$in $99.82\%$of the samples.