Question

A Gallup poll of 1,236 adults showed that 12% of the respondents believe that it is bad luck to walk under a ladder. Suppose 30 people are selected at random from among the 1,236 were polled, and further suppose we want to calculate the probability that at least 2 of the 30 believe it is bad luck to walk under a ladder. Given that the 30 subjects were selected without replacement, those selections are not independent. Can the probability be found using the binomial distribution? Why or why not? No. The selections are not independent and the 5% Rule is not satisfied. Yes. There are a fixed number of 30 selections that are independent and can be classified into two categories, and the probability of “success” does not change with each selection. No. The selections are not independent. o Yes. Although the selections are not independent, they can be treated as if they were independent by applying the 5% Rule.

Answer 1

1) Answer is

Although selections are not independent , they can be treated as if they are independent using bu applying the 5% rule.

Note : Since selections are done without replacement , each selection is dependent on the earlier selections , so they are not independent . But sample size is musch less than 5% of population size (5% of1236 = 61.8 ) , thus they can be treated as indpendent.

2) Let X follow Binomial with n= 64 , p =0.409

Then probability mass function of X is

$P(X=x)= \binom{64}{x}0.409^{x}(1-0.409)^{64-x}$, x=0,1,2,....64

To find

$P(X\leq 27)=\sum_{x=0}^{27} \binom{64}{x}0.409^{x}(1-0.409)^{64-x}$

= 0.634

3) Let X be the number of tablets which do not meet specification

X follow Binomial with n= 58 , p =0.04

To find

$P(X\leq 1)=\sum_{x=0}^{1} \binom{58}{x}0.04^{x}(1-0.04)^{58-x}$

$= \binom{58}{0}0.04^{0}(1-0.04)^{58-0}+\binom{58}{1}0.04^{1}(1-0.04)^{58-1}$

= 0.320

4) Let X be the number of voters who approve of Presidents job performance

X follow Binomial with n= 50 , p =0.39

We know that for Binomial distribution

mean = np = 50*0.39 = 19.5

variance = np(1-p) = 50*0.39*(1-0.39) = 11.9

Standard deviation = $\sqrt{np(1-p)}=3.4$

5) As n is large ( n> 50) and np = 19.5 > 10 and n(1-p) = 30.5 > 10 , we can use Normal approximation to Binomial .

Using empirical probability method , we know that 95% of the observation lie between 2 standard deviation from mean .

Outside the interval, (mean -2 SD , mean +2 SD) , values are considered unusual

Thus minimum value = mean - 2 SD

= 19.5- 2 * 3.4 = 12.7 = 13 ( nearest interger)

maximum value = mean + 2 SD

= 19.5 + 2 * 3.4 = 26.3 =26 ( nearest interger )

Out of 50 voters surveyed , 13 or fewer voters would be significantly low and 26 or more would be significantly high .