Answer:
sample size of n=33
Step by step explanation:
We can use the formula for the margin of error for a confidence interval:
Margin of error = z* (sigma / sqrt(n))
Where z* is the z-score corresponding to the desired confidence level, sigma is the population standard deviation, and n is the sample size.
In this case, we want to find the sample size n such that the margin of error is no more than 2 BPM with 95% confidence. Since the sample size is unknown, we can use a t-distribution instead of the standard normal distribution to find the appropriate critical value.
The critical value for a 95% confidence interval with n-1 degrees of freedom is t=2.064 (from a t-distribution table).
Plugging in the known values, we have:
2 = 2.064 * (11.3 / sqrt(n))
Solving for n, we get:
n = (2.064 * 11.3 / 2)^2 = 32.59
Rounding up to the nearest whole number, we need a sample size of n=33 to be within 2 BPM of the population mean with 95% confidence.