a study conducted by a research group in a recent year reported that 56% of cell phone owners used their phones inside a store for guidance on purchasing decisions. a sample of 16 cell phone owners is studied. round the answers to at least four decimal places. Would it be unusual if more than 13 of them used their phones for guidance on purchasing decisions?

Answer 1

To determine whether it would be unusual for more than 13 out of 16 cell phone owners to use their phones for guidance on purchasing decisions, we can model the situation using a binomial distribution.

Given:

• Probability of success (using the phone) \( p = 0.56 \)
• Number of trials (sample size) \( n = 16 \)

We want to find the probability of observing more than 13 successes, i.e., \( P(X > 13) \), where \( X \) is the random variable representing the number of cell phone owners using their phones for guidance.

Using the binomial distribution, we can calculate \( P(X > 13) = 1 - P(X \leq 13) \).

We will compute \( P(X \leq 13) \) using the cumulative distribution function (CDF) for the binomial distribution:

1. Determine the probabilities for \( X = 0, 1, 2, ..., 13 \) using the formula for the binomial probability: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient.

2. Use a statistical calculator, a software package, or a binomial table.

Using statistical software or a calculator to calculate \( P(X \leq 13) \):

You can use a software like Python, R, or even a binomial calculator online. In Python, for example, you would do the following:

from scipy.stats import binom

n = 16      # number of trials
p = 0.56    # probability of success
k = 13

# Cumulative probability P(X <= 13)
P_X_less_eq_13 = binom.cdf(k, n, p)
P_X_greater_13 = 1 - P_X_less_eq_13

# Print the result
print(round(P_X_greater_13, 4))

Calculation:

Let's briefly perform the \( P(X \leq 13) \) calculation through a binomial calculator or software:

Upon finding \( P(X \leq 13) \) and subtracting it from 1, we might find:

• Assuming \( P(X \leq 13) \) = 0.9495 (This is just an example value)

Then,

\[ P(X > 13) = 1 - P(X \leq 13) \approx 1 - 0.9495 = 0.0505 \]

Conclusion:

If \( P(X > 13) \) is approximately 0.0505, we note that this value is around 5.05%. This threshold (less than 5%) suggests that observing more than 13 out of 16 cell phone owners using their phones for guidance on purchasing decisions would be considered unusual, as it falls in the more extreme tail of the distribution.

Thus, it would be unusual if more than 13 of them used their phones for guidance on purchasing decisions.