Q.6.104 - Elementary Statistics

Question

Friendship Motivation. In the article "Assessing Friendship Motivation During Preadolescence and Early Adolescence" (Journal of Eurly Adolescenee, Vol. 25. No, 3. pp. 367-385). 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 to tend to score higher on the FMSC than boys. Boys in the seventh grade scored a mean of $9.32$ with a standard deviation of $1.71$, 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

a. one standard deviation to either side of the mean.

b. two standard deviations to either side of the mean.

c. three standard deviations to either side of the mean.

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

Step-by-Step Solution

Answer

a). Between $1.29kg$ and $1.51kg$.

b). Between $1.18$kg and $1.62$kg.

c). Between $1.07kg$ and $1.73kg$.

d).  Graph:

Part (a) Step 1: Given Information

Swedish men's brain weights follow a normal distribution, with $\mu =1.4kg$ and $\sigma =0.11kg$.

• The observations that lie within $1$ standard deviation from the mean are roughly $68\%$, according to the empirical for the bell-shaped distributed variable.
• Around $95\%$ of observations are within $2$ standard deviations of the mean.
• The observations that are within three standard deviations of the mean are worth $99.7\%$.

Part (a) Step 2: Explanation

The empirical rule states that the boundaries of one standard deviation from the mean are as follows:

$\mu -\sigma$$=1.29$

$\mu +\sigma$ $=1.4+0.11$

$=1.51$

As a result, around $68\%$ of brain weights fall between $1.29kg$ and $1.51kg$.

Part (b) Step 1: Given Information

Swedish men's brain weights follow a normal distribution, with $\mu =1.4kg$ and $\sigma =0.11kg$.

The observations that lie within $1$ standard deviation from the mean are roughly $68\%$, according to the empirical for the bell-shaped distributed variable.

- Around $95\%$ of observations are within two standard deviations of the mean.

- The observations that are within three standard deviations of the males are worth $99.7\%$.

Part (b) Step 2: Explanation

The borders of two standard deviations to the mean, according to the empirical rule, are:

$\mu -2\sigma$ $=1.4-2\times 0.11$

$=1.18$

$\mu +2\sigma$  $=1.4+2\times 0.11$

$=1.62$

As a result, around $95\%$ of brain weights are between $1.18kg$ and $1.62 kg$.

Part (c) Step 1: Given Information

The brain weights of Swedish men follow a normal distributionwith $\mu =1.4 \mathrm{kg}$ and $\sigma =0.11 \mathrm{kg}$.

The empirical for the bell-shaped distributed variable says that,

- The observations that lie within one standard deviation from the mean is around $68\%$.

- The observations that lie within two standard deviations from the mean is around $95\%$.

- The observations that lie within three standard deviations from the men is around $99.7\%$.

Part (c) Step 2: Explanation

The empirical rule states that the two standard deviations to the mean bounds are:

$\mu -3\sigma$ $=1.4-3\times 0.11$

$=1.07$

$\mu +3\sigma$ $=1.4+3\times 0.11$

$=1.73$

Therefore, around $99.7\%$ of brain weights are between $1.07 \mathrm{kg}$ and $1.73 \mathrm{kg}$.

Part (d) Step 1: Given Information

Swedish men's brain weights follow a normal distribution, with $\sigma =1.4kg$ and $\sigma =0.11kg$.

- The observations that lie within 1 standard deviation from the mean are roughly $68\%$, according to the empirical for the bell-shaped distributed variable.

- Around $95\%$ of observations are within two standard deviations of the mean.

- The observations that are within three standard deviations of the males are worth $99.7\%$.

Part (d) Step 2: Explanation

$X$ indicates the brain weights of Swedish men, and $z$ represents the $z$-score in the graph below.