Answer:
The sample size is [tex]n = 246[/tex]
Step-by-step explanation:
From the question we are told that
The standard deviation is [tex]\sigma = 4\ lb[/tex]
The confidence level is C = 95%
The total width is [tex]w = 1 \ lb[/tex]
Given that the confidence level is 95% then the level of significance is mathematically represented as
[tex]\alpha = (100 -95 )\%[/tex]
=> [tex]\alpha = 0.05[/tex]
From the normal distribution table the critical value for [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Generally the sample size is mathematically represented as
[tex]n = [\frac{Z_{\frac{\alpha }{2}} * \sigma }{E}]^2 [/tex]
Here E is the margin of error which is mathematically represented as
[tex]E = \frac{w}{2}[/tex]
=> [tex]E = \frac{1}{2}[/tex]
=> [tex]E = 0.5 [/tex]
So
[tex]n = [\frac{1.96 * 4 }{0.5}]^2 [/tex]
[tex]n = 246[/tex]
