Question

1. Use the given values of n and p to find the minimum usual value µ - 2s and the maximum usual value µ + 2s. Round your answer to the nearest hundredth unlessotherwise noted.

n = 267, p = 0.239 Round your answers to the nearest thousandth.

2. Use the given values of n and p to find the minimum usual value µ - 2s and the maximum usual value µ + 2s. Round your answer to the nearest hundredth unlessotherwise noted.

n = 1269, p = 0.98

Answer 1

Concepts and reason

Confidence interval:

A range of values such that the population parameter can expected to contain for the given confidence level is termed as the confidence interval. In other words, it can be defined as an interval estimate of the population parameter which is calculated for the given data based on a point estimate and for the given confidence level.

Moreover, the confidence level indicates the possibility that the confidence interval can contain the population parameter. Usually, the confidence level is denoted by . The value is chosen by the researcher. Some of the most common confidence levels are 90%, 95%, and 99%.

Binomial distribution:

In a series of n independent trials, for each trial if the probability of success is a constant p and the probability of failure is $q = 1--p$ ; then, the probability of x success and obviously $\left( {n - x} \right)$ failures is given by the Binomial distribution. Also, the binomial distribution is the discrete probability distribution.

Requirements of a binomial experiment are as follows:

• For the given experiment, each trial has two possibilities. That is, it has a success and a failure, which are mutually exclusive outcomes.

• The number of trials in the given experiment is known in advance, and they are fixed.

• The outcomes of the trials are independent.

• The probability of success is unaltered for each trail throughout the experiment.

Fundamentals

The formula for the probability of binomial distribution for the random variable X with parameters n and p is,

Where p is the probability of success on an individual trail, n is the number of trails, and x is the number of success.

The formula for mean is,

$\mu = np$

The formula for standard deviation is,

$\sigma = \sqrt {npq}$

The formula forminimum and maximum usual values is,

$\mu \pm 2s$

(1.1)

The mean value is obtained as below:

From the given information, let Xbe the random variable whichfollows binomial distribution with $n = 267$ and $p = 0.239$ . That is $X \sim {\rm{Binomial}}\left( {n = 267,p = 0.239} \right)$ .

The mean values is,

$\begin{array}{c}\\\mu = np\\\\ = 267 \times 0.239\\\\ = 63.813\\\end{array}$

Thus, the mean value is 63.813.

The standard deviation valueis obtained as follows:

The standard deviation is,

$\begin{array}{c}\\\sigma = \sqrt {267\left( {0.239} \right)\left( {1 - 0.239} \right)} \\\\ = \sqrt {63.813\left( {0.761} \right)} \\\\ = \sqrt {48.5617} \\\\ = 6.9686\\\end{array}$

Thus, the standard deviation value is 6.9686.

The minimum usual value is obtained below:

The minimum value is,

$\begin{array}{c}\\\mu - 2s = 63.813 - \left( {2 \times 6.9686} \right)\\\\ = 63.813 - 13.9372\\\\ = 49.88\\\end{array}$

(1.2)

The maximumusual value is obtained below:

The maximumvalue is,

$\begin{array}{c}\\\mu + 2s = 63.813 + \left( {2 \times 6.9686} \right)\\\\ = 63.813 + 13.9372\\\\ = 77.75\\\end{array}$

(2.1)

The mean value is obtained as below:

From the given information, let Xbe the random variable whichfollows binomial distribution with $n = 1,269$ and $p = 0.98$ . That is $X \sim {\rm{Binomial}}\left( {n = 1,269,p = 0.98} \right)$ .

The mean values is,

$\begin{array}{c}\\\mu = np\\\\ = 1,269 \times 0.98\\\\ = 1,243.62\\\end{array}$

Thus, the mean value is 1,243.62.

The standard deviation valueis obtained as follows:

The standard deviation is,

$\begin{array}{c}\\\sigma = \sqrt {1,269\left( {0.98} \right)\left( {1 - 0.98} \right)} \\\\ = \sqrt {1,243.62\left( {0.02} \right)} \\\\ = \sqrt {24.8724} \\\\ = 2.2332\\\end{array}$

Thus, the standard deviation value is 2.2332.

The minimum usual value is obtained below:

The minimum value is,

$\begin{array}{c}\\\mu - 2s = 1,243.62 - \left( {2 \times 2.2332} \right)\\\\ = 1,243.62 - 4.4664\\\\ = 1,239.15\\\end{array}$

(2.2)

The maximum usual value is obtained below:

The maximumvalue is,

$\begin{array}{c}\\\mu + 2s = 1,243.62 + \left( {2 \times 2.2332} \right)\\\\ = 1,243.62 + 4.4664\\\\ = 1,248.09\\\end{array}$

Ans: Part 1.1

Thus, the minimum usual value is 49.88.

Part 1.2

Thus, the maximum usual value is 77.75.

Part 2.1

Thus, the minimum usual value is 1,239.15.

Part 2.2

Thus, the maximum usual value is 1,248.09.