Answer to Question #244916 in Statistics and Probability for hetr

Question

Of the population of customers of a store, a proportion of 0.4 are not satisfied with the service offered by the store. In a random sample of 400 customers, what is the probability that between 156 and 240 customers will be satisfied with the service offered by the store?

Answer 1

The situation can be described using Binomial distribution Y~Bin(400, 0.6).

Since there is large sample size(400 elements), then, according to the central limit theorem, ${\displaystyle \mathrm {Bin} (n,p)\approx N(np,npq)}$, where N is normal distribution.

In the given case:

${\displaystyle \mathrm {Bin} (400,0.6)\approx N(400*0.6,400*0.6*0.4)}= N(240, 96) = 240+9.8Z$, where Z is the standard normal distribution

So, Y~$240+9.8Z$

P(156<240+9.8Z<240) = P(-8.57<Z<0)= P(Z<0) - P(Z<-8.57) $\approx 0.5$ (the value of P(Z<-8.57) is very close to 0)

So, the seeking probability is insignificantly less than 0.5.