Answer to Question #269792 in Statistics and Probability for Mayyy

Question

In the Binomlal Distribution, there is a fraction of 0.014 of the defective production output. Only 95 items were taken as samples. Asumi items - defective items occur randomly and freely. a). Determine the value of the standard deviation of the proportion of defects in this number of samples b). What is the probability that the defective items in the sample are in the range of values ​​2 times the standard deviation of the defective fraction of the entire population?

Answer 1

a) $\sigma=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.014((1-0.014)}{95}}=0.0121.$

b) $P(\mu-2\sigma<X<\mu+2\sigma)=P(-2<Z<2)=0.9544.$