To solve this problem, we need to use the normal distribution since we are dealing with sample proportions. We can calculate the mean and standard deviation of the sample proportion as follows:
Mean = p = 0.4
Standard deviation = sqrt[(p(1-p))/n] = sqrt[(0.4*0.6)/200] = 0.03464
To find the probability that the sample proportion will be between 40 and 43, we need to find the z-scores corresponding to those values and use the normal distribution.
z1 = (0.4 - 0.4)/0.03464 = 0
z2 = (0.43 - 0.4)/0.03464 = 0.8674
Using a z-table or calculator, we can find that the probability of a z-score between 0 and 0.8674 is 0.3065. Therefore, the probability that the sample proportion of depositors with multiple accounts will be between 40 and 43 is 0.3065.
The manager of the local branch of saving bank has determined that 40% of all depositors have multiple accounts at the bank if a rondum sample of 200 depositors is selected what is the probability that the sample proportion of depositors with multiple accounts will be between 40 and 43
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