Question

Vitamin D is essential for strong, healthy bones. Our bodies produce vitamin D naturally when sunlight falls upon the skin, or it can be taken as a dietary supplement. Although the bone disease rickets was largely eliminated in England during the 1950 s, some people there are concerned that this generation of children is at increased risk because they are more likely to watch TV or play computer games than spend time outdoors. Recent research indicated that about \(20 \%\) of British children are deficient in vitamin D. Suppose doctors test a group of elementary school children. a) What's the probability that the first vitamin Ddeficient child is the 8 th one tested? b) What's the probability that the first 10 children tested are all okay? c) How many kids do they expect to test before finding one who has this vitamin deficiency? d) They will test 50 students at the third-grade level. Find the mean and standard deviation of the number who may be deficient in vitamin D. e) If they test 320 children at this school, what's the probability that no more than 50 of them have the vitamin deficiency?

Short answer

a) 0.0328; b) 0.107; c) 5; d) Mean 10, SD 2.83; e) ≈0.025.

Step-by-step solution

Identify Type of Probability Distribution

For parts (a), (b), and (c), recognize that we are dealing with a geometric distribution since we are looking for the first occurrence of an event (vitamin D deficiency). For parts (d) and (e), identify that the problem moves towards a binomial distribution since we are concerned with a specific number of occurrences within a fixed number of trials.

Probability That First Deficient Child is the 8th One Tested

Use the geometric distribution formula, which is given by: \[ P(X = k) = (1-p)^{k-1} \cdot p \]where \( p = 0.2 \) is the probability of a child being vitamin D deficient and \( k = 8 \). Substitute values:\[ P(X = 8) = (0.8)^{7} \cdot 0.2 \approx 0.0328 \]

Probability That First 10 Children Are Okay

Here, we want the probability that none of the first 10 children are deficient, which uses:\[ P(X > 10) = (1-p)^{10} = 0.8^{10} \approx 0.107 \]

Expected Number of Children Tested to Find First Deficient One

For geometric distribution, the expected value \( \text{E}(X) \) is given by:\[ \text{E}(X) = \frac{1}{p} = \frac{1}{0.2} = 5 \]

Mean and Standard Deviation for 50 Tests

Since this is a binomial distribution with \( n = 50 \) and \( p = 0.2 \), calculate the mean and standard deviation as:\[ \text{Mean} = n \cdot p = 50 \cdot 0.2 = 10 \]\[ \text{Standard Deviation} = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{50 \cdot 0.2 \cdot 0.8} = \sqrt{8} \approx 2.83 \]

Probability No More than 50 Out of 320 Deficient

Since this involves a large sample size, use the normal approximation for the binomial distribution here, with mean \( \mu = n \cdot p = 320 \cdot 0.2 = 64 \) and variance \( \sigma^2 = 320 \cdot 0.2 \cdot 0.8 = 51.2 \), so \( \sigma \approx 7.16 \).Using the normal approximation: First, find the Z-score: \[ Z = \frac{X - \mu}{\sigma} = \frac{50 - 64}{7.16} \approx -1.96 \]Now, look up \( Z = -1.96 \) in normal distribution tables, or use software, which gives roughly 0.025.

Key concepts

Geometric Distribution

The geometric distribution is quite useful when we want to find the probability of the first occurrence of an event after a certain number of trials. In our exercise, we are interested in the first vitamin D deficient child appearing in the group of kids being tested. The key feature of a geometric distribution is that each child tested is independent of the others. Each child has the same probability of having a vitamin D deficiency, denoted by \( p = 0.2 \). This means there's an 80% chance a child does not have the deficiency, or \( 1 - p = 0.8 \).

In part a) of the exercise, we calculate the probability that it takes exactly 8 children for the first deficient one to appear. The formula \( P(X = k) = (1-p)^{k-1} \cdot p \) helps us arrive at the answer. It involves multiplying the probability all prior trials (the first 7 children) did not result in a deficiency, by the probability the 8th child does.

For part b), the task reverses to find the probability that no child among the first 10 tested has a deficiency. Here, we simply use \( P(X > 10) = (1-p)^{10} \), which accounts for all children not being deficient.
Finally, part c) involves finding the expected number of kids tested until the first deficiency, calculated by \( \text{E}(X) = \frac{1}{p} \), producing 5. This means we expect to test 5 children on average before finding one with the deficiency.

Binomial Distribution

The binomial distribution is an extension of the concept behind individual trials, applied when we have a fixed number of trials to consider at once. In our scenario, each child is a trial, and we have multiple trials representing all children we're testing for vitamin D deficiency. The probability of deficiency, \( p \), remains at 0.2.

In part d), we are asked about expected outcomes over 50 trials. The mean or expected number of deficient children is given by \( \text{Mean} = n \cdot p \), where \( n = 50 \). This calculation tells us we can expect 10 deficient children out of the 50 tested. Similarly, the calculation for standard deviation signals typical deviation in these results. Using the formula \( \sqrt{n \cdot p \cdot (1-p)} \), we find a standard deviation of approximately 2.83, indicating the spread of our anticipated results around the mean.

Unlike the geometric distribution that focuses on when the first success occurs, the binomial distribution assesses the quantity of successes over a fixed number of trials, with each trial being independent.

Normal Approximation

In cases where the sample size is large, such as testing 320 children, it becomes sensible to use the normal approximation to a binomial distribution. This is because evaluating probabilities directly from the binomial distribution can become cumbersome as the number counts grow.

The normal approximation transforms the binomial into a continuous normal distribution, making it simpler to compute probabilities for ranges of outcomes. This is possible when the sample size \( n \) is large and probabilities are not too close to 0 or 1.

For our exercise's part e), we first calculate the mean \( \mu = n \cdot p \), which is 64 in this case, and standard deviation \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \), approximately 7.16. Using these, we convert the problem to finding the Z-score \( Z = \frac{X - \mu}{\sigma} \), and then consult standard normal distribution tables for the probability of no more than 50 children being deficient.
The normal distribution making computations easier is why it's commonly used as an approximation in real-world situations, especially for large sample sizes.