Chapter 6.3, Problem 99E

(a)

To determine

To explain: L can be modelled by a binomial distribution.

Explanation of Solution

Given information:

L is the number of left handed students in the sample.

Sample size = 100

Percentage of left handed students at a large school = 11%

Concept used:

10% condition

n < 0.10N

The rule says that if the sample is less than $10\%$ of the population, then it is reasonable to assume that the trials are independent and can be modelled by binomial distribution irrespective of the without replacement sample.

Here, the sample size ( $=100$ ) is less than $10\%$ of students at a large school.

Moreover, here the left handedness of one student does not give any idea of the left handedness of other student, so trial of each student is independent.

So, $\text{L }~\text{ Bin }\left(\text{1}00,0.\text{11}\right)$

Hence, L can be modelled by a binomial distribution.

(b)

To determine

To calculate: Probability that more than or equal to 15 students in the sample are left handed.

Answer to Problem 99E

Probability that more than or equal to 15 students in the sample are left handed is 0.13305.

Explanation of Solution

Calculation:

  $\begin{array}{l}\text{P }(\text{L }\ge \text{ 15) = 1 - P (L < 15)}\\ \text{P }(\text{L }\ge \text{ 15) = 1 - P (L }\le \text{ 14)}\end{array}$

Using TI- 83 plus

a) Click on ${{\displaystyle 2}}^{nd}$ and then click on Dist .

b) Go to binomcdf with $\downarrow$ key.

c) Hit Enter

d) Type 100, 0.11, 14)

e) Hit Enter

The probability comes to be 0.86695.

  

  $\begin{array}{l}\text{P }\left(\text{L }\ge \text{ 15}\right)\text{ = 1 - 0}\text{.86695}\\ \text{P }\left(\text{L }\ge \text{ 15}\right)\text{ = }0.13305\end{array}$