Q. 6.104 - Elementary Statistics

Question

In the article "Assessing Friendship Motivation During Preadolescence and Early Adolescence", J. Richard and B. Schneider described the properties of the Friendship Motivation Scale for Children (FMSC), a scale designed to assess children's desire for friendships. Two interesting conclusions are that friends generally report similar levels of the FMSC and girls tend to score higher on the FMSC than boys. Boys in the seventh grade scored a mean $9.32$ with a standard deviation of data-custom-editor="chemistry" $1.71,$ and girls in the seventh grade scored a mean of $10.04$ with a standard deviation of $1.83.$ Assuming that FMSC scores are normally distributed, determine the percentage of seventh-grade boys who have FMSC scores within

Part (a): One standard deviation to either side of the mean.

Part (b): Two standard deviation to either side of the mean.

Part (c): Three standard deviation to either side of the mean.

Part (d): Repeat parts (a)-(c) for seventh-grade girls.

Step-by-Step Solution

Answer

Part (a): The percentage of seventh-grade boys who have FMSC scores within one standard deviation to either side of the mean is $7.61$ and $11.03$.

Part (b): The percentage of seventh-grade boys who have FMSC scores within two standard deviation to either side of the mean is $5.9$ and $12.74$.

Part (c): The percentage of seventh-grade boys who have FMSC scores within three standard deviation to either side of the mean is $4.19$ and $14.45$.

Part (d): The percentage of seventh-grade girls who have FMSC scores within one standard deviation to either side of the mean is $8.21$ and $11.87$.

The percentage of seventh-grade girls who have FMSC scores within two standard deviation to either side of the mean is $6.38$ and $13.7$.

The percentage of seventh-grade girls who have FMSC scores within three standard deviation to either side of the mean is $4.55$ and $15.53$.

Part (a) Step 1. Given information.

The given mean of boys is $9.32$ and standard deviation is $1.71$.

The given mean of girls is $10.04$ and standard deviation is $1.83$.

Part (a) Step 2. Find the one standard deviation above and below mean of boys.

Property $1$ of the $68.26-95.44-99.74$rule says $68.26\%$ of all FMSC scores of boys have duration within one standard deviation to either side of the mean.

One standard deviation below mean is given below,

$$\begin{gathered} \mu -\sigma =9.32-1.71 \\ =7.61 \end{gathered}$$

One standard deviation above mean is given below,

$$\begin{gathered} \mu +\sigma =9.32+1.71 \\ =11.03 \end{gathered}$$

Therefore, $68.26\%$ of all FMSC scores of boys are between $7.61,11.03$.

Part (b) Step 1. Find the two standard deviation above and below mean of boys.

Property $2$ of the $68.26-95.44-99.74$ rule says $95.4\%$ of all FMSC scores of boys have duration within two standard deviation to either side of the mean.

Two standard deviation below mean is given below,

$$\begin{gathered} \mu -\sigma 2=9.32-\left(2\right)1.71 \\ =5.9 \end{gathered}$$

Two standard deviation above mean is given below,

$$\begin{gathered} \mu +\sigma 2=9.32+\left(2\right)1.71 \\ =12.74 \end{gathered}$$

Therefore,$95.4\%$ of all FMSC scores of boys are between $5.9,12.74$.

Part (c) Step 1. Find the three standard deviation above and below mean of boys.

Property $3$ of the $68.26-95.44-99.74$ rule says $99.74\%$ of all FMSC scores of boys have duration within three standard deviations to either side of the mean.

Three standard deviation below mean is given below,

$$\begin{gathered} \mu -3\sigma =9.32-\left(3\right)1.71 \\ =4.19 \end{gathered}$$

Three standard deviation above mean is given below,

$$\begin{gathered} \mu +3\sigma =9.32+\left(3\right)1.71 \\ =14.45 \end{gathered}$$

Therefore, $99.74\%$of all FMSC scores of boys are between $4.19,14.45$.

Part (d) Step 1. Find the one standard deviation above and below mean of girls.

Property $1$ of the $68.26-95.44-99.74$ rule says $68.26\%$ of all FMSC scores of girls have duration within one standard deviation to either side of the mean.

One standard deviation below mean is given below,

$$\begin{gathered} \mu -\sigma =10.04-1.83 \\ =8.21 \end{gathered}$$

One standard deviation above mean is given below,

$$\begin{gathered} \mu +\sigma =10.04+1.83 \\ =11.87 \end{gathered}$$

$68.26\%$of all FMSC scores of girls are between $8.21,11.87$.

Part (d) Step 2. Find the two standard deviation above and below mean of girls.

Property $2$ of the $68.26-95.44-99.74$ rule says $95.4\%$of all FMSC scores of girls have duration within two standard deviation to either side of the mean.

Two standard deviation below mean is given below,

$$\begin{gathered} \mu -2\sigma =10.04-2\left(1.83\right) \\ =6.38 \end{gathered}$$

Two standard deviation above mean is given below,

$$\begin{gathered} \mu +2\sigma =10.04+2\left(1.83\right) \\ =13.7 \end{gathered}$$

Therefore, $95.4\%$of all FMSC scores of girls are between $6.38,13.7$.

Part (d) Step 3. Find the three standard deviation above and below mean of girls.

Property $3$ of the $68.26-95.44-99.74$ rule says $99.74\%$ of all FMSC scores of girls have duration within three standard deviations to either side of the mean.

Three standard deviation below mean is given below,

$$\begin{gathered} \mu -3\sigma =10.04-\left(3\right)1.83 \\ =4.55 \end{gathered}$$

Three standard deviation above mean is given below,

$$\begin{gathered} \mu +3\sigma =10.04+\left(3\right)1.83 \\ =15.53 \end{gathered}$$

Therefore, $99.74\%$ of all FMSC scores of girls are between $4.55,15.53$.