Question

90-day forecast
weather type | predicted frequency | observed frequency
wind | 9 | 12
rain | 18 | 15
clouds | 27 | 33
sunshine | 38 | 30

the table shows the results of the experiment to determine the 90-day weather forecast. how does the experimental probability compare to the theoretical probability? the theoretical probability for rain is 1/5. the experimental probability for rain is blank, the actual weather is blank the theoretical probability.

Explanation:

Step1: Calculate total days

The total number of days in the experiment is the sum of all observed (or predicted) frequencies. Using observed: \(12 + 15 + 33 + 30 = 90\).

Step2: Find experimental probability of rain

Experimental probability of rain is \(\frac{\text{Observed frequency of rain}}{\text{Total number of days}}=\frac{15}{90}=\frac{1}{6}\approx0.1667\).

Step3: Compare with theoretical probability

Theoretical probability of rain is \(\frac{1}{5}=0.2\). Convert to common denominator: \(\frac{1}{6}=\frac{5}{30}\), \(\frac{1}{5}=\frac{6}{30}\). Since \(\frac{5}{30}<\frac{6}{30}\), experimental probability is less than theoretical.

Answer:

First, we need to calculate the experimental probability of rain. The total number of days is \(9 + 18 + 27 + 36=90\) (from the predicted frequencies, we can also check with observed: \(12 + 15 + 33 + 30 = 90\)). The observed frequency of rain is \(15\). So the experimental probability of rain is \(\frac{15}{90}=\frac{1}{6}\approx0.1667\). The theoretical probability is given as \(\frac{1}{5} = 0.2\). Now we compare \(\frac{1}{6}\) and \(\frac{1}{5}\). To compare, we can find a common denominator: \(\frac{1}{6}=\frac{5}{30}\), \(\frac{1}{5}=\frac{6}{30}\). Since \(\frac{5}{30}<\frac{6}{30}\), the experimental probability (\(\frac{1}{6}\)) is less than the theoretical probability (\(\frac{1}{5}\)).