Question

Psychiatrists estimate that 1 in 100 adults suffers from bipolar disorder. Use the normal approximation to the binomial distribution to estimate the probability that in a city of 10,000 people there are more than 120 people who are bipolar.

Answer 1

To estimate the probability that in a city of 10,000 people there are more than 120 people who are bipolar, we can use the normal approximation to the binomial distribution. The probability is approximately 2.28%.

Step-by-step explanation:

To estimate the probability that in a city of 10,000 people there are more than 120 people who are bipolar, we can use the normal approximation to the binomial distribution. The first step is to calculate the mean and standard deviation of the binomial distribution. The mean (μ) is given by μ = n * p, where n is the number of trials (10,000) and p is the probability of success (1 in 100 or 0.01). In this case, μ = 10,000 * 0.01 = 100. The standard deviation (σ) is given by σ = sqrt(n * p * (1 - p)). In this case, σ = sqrt(10,000 * 0.01 * (1 - 0.01)) = sqrt(100) = 10.

Next, we need to use the normal distribution to calculate the probability. To do this, we standardize the number of successes (120) using the z-score formula: z = (x - μ) / σ. In this case, z = (120 - 100) / 10 = 2.

Finally, we can use a standard normal distribution table or calculator to find the probability that a z-score is greater than 2. The probability is approximately 0.0228, or 2.28%.

Answer 2

0.0222 is the probability that in a city of 10,000 people there are more than 120 people who are bipolar.

Step-by-step explanation:

We are given the following in the question:

Probability of adult suffering from bipolar disorder =

`p =(1)/(100) = 0.01`

Sample size, n = 10000

We have to use normal approximation to the binomial distribution to estimate the probability.

Normal approximation:

`\mu = np = 10000(0.01) = 10\\\sigma = √(np(1-p)) = √(10000(0.01)(1-0.01)) = 9.95`

The distribution of adults suffering from bipolar disorder follows a normal distribution with mean 100 and standard deviation 9.5

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

We have to evaluate:

P(more than 120 people are bipolar)

P(x > 120)

`P( x > 120) = P( z > \displaystyle(120 - 100)/(9.95)) = P(z > 2.010)`

`= 1 - P(z \leq 2.010)`

Calculation the value from standard normal z table, we have,

`P(x > 610) = 1 - 0.9778 = 0.0222 = 2.22\%`

0.0222 is the probability that in a city of 10,000 people there are more than 120 people who are bipolar.