Question 1177608: The times spent by people visiting a certain dentist are independent and normally distributed with a mean of 8.2 minutes. 79% of people who visit this dentist have visits lasting less than 10 minutes.
Find the probability that, of 35 randomly chosen people, fewer than 16 have visits lasting
less than 8.2 minutes.
(answer = 0.250)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **Step 1: Find the standard deviation**
We know that the visit times are normally distributed with a mean (µ) of 8.2 minutes. We also know that 79% of the visits last less than 10 minutes. This information allows us to find the standard deviation (σ) of the distribution.
* Let X be the random variable representing the time spent by a person visiting the dentist.
* We are given P(X < 10) = 0.79.
* We can standardize this value by finding the z-score corresponding to a cumulative probability of 0.79. Using a standard normal table or calculator, we find that the z-score is approximately 0.81.
Now, using the z-score formula:
```
z = (x - µ) / σ
```
We can plug in the values:
```
0.81 = (10 - 8.2) / σ
```
Solving for σ:
```
σ ≈ 2.22 minutes
```
**Step 2: Find the probability of a visit lasting less than 8.2 minutes**
Since the mean is 8.2 minutes, the probability of a visit lasting less than 8.2 minutes is simply 0.5 (because the normal distribution is symmetric around the mean).
**Step 3: Find the probability of fewer than 16 people out of 35 having visits less than 8.2 minutes**
Now we have a binomial distribution problem.
* n = 35 (number of trials)
* p = 0.5 (probability of success, i.e., a visit lasting less than 8.2 minutes)
* We want to find P(X < 16), where X is the number of people with visits less than 8.2 minutes.
We can use the binomial probability formula or a binomial calculator to find this probability. Using a calculator, we get:
```
P(X < 16) ≈ 0.214
```
**Therefore, the probability that fewer than 16 out of 35 randomly chosen people have visits lasting less than 8.2 minutes is approximately 0.214.**
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