Final answer:
By using normal approximation to the binomial distribution, we find that there is approximately a 75.49% chance that more than 107 out of 680 smartphone users surveyed use Apple smartphones.
Step-by-step explanation:
To calculate the probability of more than 107 Apple smartphone users in a survey of 680, knowing the market share is 0.1657, we can use the normal approximation to the binomial distribution because the sample size is large. The mean (μ) of the distribution is n*p, and the standard deviation (σ) is sqrt(n*p*(1-p)), where 'n' is the sample size and 'p' is the proportion of success (Apple users).
First, we calculate the mean and standard deviation:
- μ = n * p = 680 * 0.1657 = 112.676
- σ = sqrt(n * p * (1 - p)) = sqrt(680 * 0.1657 * (1 - 0.1657)) ≈ sqrt(680 * 0.1657 * 0.8343) ≈ 8.24
Next, we convert the value of interest (107) to a z-score:
z = (X - μ) / σ = (107 - 112.676) / 8.24 ≈ -0.69
Since we are looking for the probability of having more than 107 users, we will need the complementary cumulative probability associated with the z-score of -0.69, which can be found using a standard normal distribution table or a calculator.
We are looking for P(X > 107), which corresponds to 1 - P(Z < -0.69).
Assuming the calculator shows P(Z < -0.69) ≈ 0.2451, the probability that more than 107 out of 680 smartphone users have an Apple smartphone is:
P(X > 107) = 1 - 0.2451 = 0.7549 or 75.49%.