Chapter PIV, Problem 13RE

(a)

To determine

To find: the probability that no greater than fifteen of them are in control of left-brain language.

Answer to Problem 13RE

0.17

Explanation of Solution

Given:

  $\begin{array}{c}P\left(L\right)=0.8\\ P\left(R\right)=0.1\\ P\left(L\cap R\right)=0.1\end{array}$

Calculation:

Suppose

L= Language skills reside on the left side of the brain.

R = Language skills are on the right side of the brain.

Let X be the number of individuals with control of left-brain language.

  $P\left(L\right)=0.8$

X is binom (25,0.8)

Probability that no more than 15 have left brain language control

  $\begin{array}{c}P\left(X\le 15\right)={\displaystyle \sum _{x=0}^{15}C{}_x{\left(0.8\right)}^x{\left(0.2\right)}^{25-x}}\\ =0.01733\end{array}$

There is a possibility of 1.7 percent that 15 have left brain language

(b)

To determine

To find: the probability of no one having two-sided control of language.

Answer to Problem 13RE

0.59

Explanation of Solution

Given:

  $\begin{array}{c}n=5\\ r=0\end{array}$

Calculation:

Let X be number of students with two-sided language control.

X is binom (5, 0.1)

Probability that no one has two-sided language control

  $\begin{array}{c}P\left(X=0\right)=C{}_0{\left(0.1\right)}^0{\left(0.9\right)}^5\\ =0.59\end{array}$

There is 59% possibility that no one has two-sided language control.

(c)

To determine

To find: the amount that of type would expect to find.

Explanation of Solution

1200 is the number of students in a class.

The expected number of left-sided language control students is

  = 1200(0.8) =960 

The expected number of students with language control on the right side is

1200(0.1) = 120

The expected number of double-sided language control students is

1200(0.1) = 120

(d)

To determine

To find: The mean and standard deviation of the number of such freshmen who can be right-brained in language skills can be found.

Answer to Problem 13RE

Mean = 120

Standard Deviation = 10.39

Explanation of Solution

Given:

  $\begin{array}{c}n=1200\\ p=0.1\end{array}$

Formula used:

  $\begin{array}{c}\mu =np\\ \sigma =\sqrt{npq}\end{array}$

Calculation:

Mean number of students with Right-brained in language ability is

1200(0.1)

  =120

Standard deviation is

  $\sqrt{npq}=\sqrt{1200\left(0.1\right)\left(0.9\right)}=10.39$

(e)

To determine

To find out: there could be accurate brain language control for the number of students in the freshman class.

Explanation of Solution

Using the rule

(for 1 standard deviation) The number of students with right regulation of brain language is

  $\begin{array}{c}\left(\mu -\sigma ,\mu -\sigma \right)=\left[120-10.39,120+10.39\right]\\ =\left[109.61,130.39\right]\end{array}$

(for 2 standard deviation) The number of students with right regulation of brain language is

  $\begin{array}{c}\left(\mu -2\sigma ,\mu -2\sigma \right)=\left[120-2\left(10.39\right),120+2\left(10.39\right)\right]\\ =\left[99.22,140.78\right]\end{array}$

(for 3 standard deviation) The number of students with correct regulation of brain language is

  $\begin{array}{c}\left(\mu -3\sigma ,\mu -3\sigma \right)=\left[120-3\left(10.39\right),120+3\left(10.39\right)\right]\\ =\left[88.83,151.17\right]\end{array}$