Answer:
a
mean [tex]\mu = 10[/tex] variance [tex]\sigma^2 = 6[/tex]
b
The binomial random variable x fall into this interval ranges from
- 5 to 5
c
[tex]P(6 \le x \le 14) = 0.8969[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 25[/tex]
The percentage that look for gas stations and food outlets that are close to or visible from the highway is [tex]p = 0.40[/tex]
Generally the mean is mathematically represented as
[tex]\mu = n * p[/tex]
=> [tex]\mu = 0.40 * 25[/tex]
=> [tex]\mu = 10[/tex]
The variance is mathematically represented as
[tex]\sigma^2 = np(1- p )[/tex]
=> [tex]\sigma^2 = 25 * 0.40(1- 0.40 )[/tex]
=> [tex]\sigma^2 = 6[/tex]
The standard deviation is mathematically evaluated as
[tex]\sigma = \sqrt{\sigma^2}[/tex]
[tex]\sigma = \sqrt{6}[/tex]
[tex]\sigma = 2.45[/tex]
The interval is evaluated as
[tex]p\pm 2 \sigma[/tex]
=> [tex]p - 2 \sigma\ \ \ , \ \ \ p + 2\sigma[/tex]
=> [tex]0.40 - 2 *2.45\ \ \ , \ \ \ 0.40 + 2* 2.45[/tex]
=> [tex]-4.5\ \ \ , \ \ \ 5.3[/tex]
The binomial random variable x fall into this interval ranges from
- 5 to 5
Generally
[tex]P(6 \le x \le 14) = P(\frac{ x - \mu }{\sigma } \le \frac{14 - 10}{{2.45}} ]-P[ \frac{ x - \mu }{\sigma } \le \frac{6 - 10}{2.45 } ][/tex]
[tex]P(6 \le x \le 14) = P(Z \le 1.63 ]-P[ Z \le -1.63 ][/tex]
[tex]P(6 \le x \le 14) = [1- P(Z > 1.63 ]] -[1- P[ Z > -1.63 ]][/tex]
From the z-table
[tex]P(Z > 1.63 ) = 0.051551[/tex]
And
[tex]P(Z >- 1.63 ) =0.94845[/tex]
=> [tex]P(6 \le x \le 14) = [1-0.051551] -[1-0.94845][/tex]
=> [tex]P(6 \le x \le 14) = 0.8969[/tex]
