Question

4. As part of a promotion for a new type of cracker, free samples are offered to shoppers in a local supermarket. The probability that a shopper will buy a packet of crackers after tasting the free sample is 0.3. Different shoppers can be regarded as independent trials. Let p be the proportion of the next 60 shoppers that buy a packet of the crackers after tasting a free sample. a) What are the mean and standard deviation of the distribution of the sample proportion

Answer 1

Answer:

For this case we need to check the conditions in order to use the normal approximation:

1) $np = 60*0.3= 18>10$

2) $n(1-p) = 60*(1-0.3)= 42>10$

Since both conditions are satisfied and the independence condition is assumed we can use the normal approximation given by:

$\hat p \sim N (\hat p, \sqrt{\frac{\hat p(1-\hat p)}{n}})$

The mean would be given by:

$\mu_{\hat p}=0.3$

And the deviation is given by:

$\sigma_{\hat p}= \sqrt{\frac{0.3*(1-0.3)}{60}}= 0.0592$

Step-by-step explanation:

For this case we know the following info:

n =60 represent the sample size

$\hat p = 0.3$ represent the estimated proportion of people that will buy a packet of crackers after tasting

For this case we need to check the conditions in order to use the normal approximation:

1) $np = 60*0.3= 18>10$

2) $n(1-p) = 60*(1-0.3)= 42>10$

Since both conditions are satisfied and the independence condition is assumed we can use the normal approximation given by:

$\hat p \sim N (\hat p, \sqrt{\frac{\hat p(1-\hat p)}{n}})$

The mean would be given by:

$\mu_{\hat p}=0.3$

And the deviation is given by:

$\sigma_{\hat p}= \sqrt{\frac{0.3*(1-0.3)}{60}}= 0.0592$