Question

An article claimed that only 38% of hotel visitors used the alarm clock provided. A hotel manager wanted to see whether that proportion applied to the hotel where she worked, so she took a random sample of 700 and checked whether they used their room alarm clock. Of the sampled visitors, only 35% used the alarm clock.

What is the approximate probability that less than 35% of the sample used the alarm clock (rounded to the nearest hundreth)?

Answer 1

The approximate probability that less than 35% of hotel visitors in the sample used the alarm clock is 4.75%. This is calculated using the normal distribution approximation for the binomial distribution, which involves finding the standard error and converting the observed proportion to a z-score.

Step-by-step explanation:

The question asks for the probability that less than 35% of the sample used the alarm clock. To approximate this probability, we use the normal distribution as an approximation to the binomial distribution because the sample size is large (n=700). First, we need to calculate the standard error (SE) using the following formula:

SE = sqrt(p * (1 - p) / n)

Where:

p is the proportion given by the article, 0.38.

n is the sample size, 700.

SE = sqrt(0.38 * (1 - 0.38) / 700) = sqrt(0.38 * 0.62 / 700) = sqrt(0.2356 / 700) = sqrt(0.000336571) ≈ 0.018

Next, we will calculate the z-score for the observed proportion in the sample (0.35):

z = (observed proportion - expected proportion) / SE

z = (0.35 - 0.38) / 0.018 ≈ -1.67

Using a z-table or a standard normal distribution calculator, we can find the probability that corresponds to the z-score of -1.67. The probability is approximately 0.0475, or 4.75% when converted to a percentage.

Therefore, the approximate probability that less than 35% of the sample used the alarm clock is 4.75%, rounded to the nearest hundredth as requested.