An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via "smart phones", so they want to estimate the proportion of users who access the site that way(even if they also use their computers sometimes). They draw a random sample of 100 investors from their customers. Suppose that the true proportion of smart phone users is 33%. a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users be?
b) What is the probability that the sample proportion of smart phone users is greater than
0.33?
c) What is the probability that the sample proportion is between 0.19 and 0.31?

a) 0.047

b) 50% probability that the sample proportion of smart phone users is greater than 0.33.

c) 33.39% probability that the sample proportion is between 0.19 and 0.31

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean and standard deviation

In this question, we have that:

a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users be?

b) What is the probability that the sample proportion of smart phone users is greater than 0.33?

This is 1 subtracted by the pvalue of Z when X = 0.33. So

By the Central Limit Theorem

has a pvalue of 0.5

1 - 0.5 = 0.5

50% probability that the sample proportion of smart phone users is greater than 0.33.

c) What is the probability that the sample proportion is between 0.19 and 0.31?

This is the pvalue of Z when X = 0.31 subtracted by the pvalue of Z when X = 0.19. So

X = 0.31

has a pvalue of 0.3354

X = 0.19

has a pvalue of 0.0015

0.3354 - 0.0015 = 0.3339

33.39% probability that the sample proportion is between 0.19 and 0.31

translate k to n and reflect across the line containing jk.