Let X and Y equal the respective numbers of hours a randomly selected child watches movies or cartoons on TV during a certain month. From experience, it is known that E(X) = 30, E(Y) = 50, Var(X) = 52, Var(Y) = 64, and Cov(X, Y)= 14. Twenty-five children are selected at random. Let Z equal the total number of hours these 25 children watch TV movies or cartoons in the next month. Approximate P(1970 < Z < 2090)
To approximate P(1970 < Z < 2090), we need to use properties of the sum of random variables.
Let's denote the total number of hours watched by the 25 children as Z. We know that E(Z) = E(25X) + E(25Y) = 25E(X) + 25E(Y) = 25 * 30 + 25 * 50 = 2000.
The variance of the sum of independent random variables is the sum of their variances. Since the children's TV watching behaviors are assumed to be independent, we have Var(Z) = Var(25X) + Var(25Y) = 25^2 * Var(X) + 25^2 * Var(Y) = 25^2 * 52 + 25^2 * 64 = 162500.
The covariance of the sum of independent random variables is the sum of their covariances. Since the children's TV watching behaviors are assumed to be independent, Cov(Z) = Cov(25X, 25Y) = 25^2 * Cov(X, Y) = 25^2 * 14 = 8750.
To approximate P(1970 < Z < 2090), we can use the normal approximation to the distribution of Z. Z can be approximated as a normal random variable with mean E(Z) = 2000 and standard deviation sqrt(Var(Z)) = sqrt(162500) = 403.69.
To calculate the z-scores for the lower and upper bounds, we use the formula z = (x - mean) / standard deviation.
For 1970: z1 = (1970 - 2000) / 403.69 = -0.743.
For 2090: z2 = (2090 - 2000) / 403.69 = 2.217.
Next, we look up the cumulative probabilities for these z-scores in the standard normal distribution table. Pr(-0.743 < Z < 2.217) is the difference between these cumulative probabilities.
Using a standard normal distribution table or a calculator that can compute the cumulative probability, we find that Pr(-0.743 < Z < 2.217) is approximately 0.782.
Therefore, the approximate probability P(1970 < Z < 2090) is approximately 0.782.
To approximate the probability, we need to use the concept of the Central Limit Theorem (CLT) since we are dealing with a large sample size of 25 children.
According to the CLT, the sum of a large number of independent and identically distributed random variables will be approximately normally distributed, regardless of the distribution of the individual variables.
Let's calculate the mean and variance of Z:
Mean of Z = E(Z) = 25 * E(X + Y) = 25 * (E(X) + E(Y)) = 25 * (30 + 50) = 2000
Variance of Z = Var(Z) = 25 * (Var(X) + Var(Y) + 2 * Cov(X, Y)) = 25 * (52 + 64 + 2 * 14) = 2600
Using the mean and variance, we can calculate the standard deviation of Z:
Standard deviation of Z = sqrt(Variance of Z) = sqrt(2600) ≈ 50.99
Now, we need to standardize the values of 1970 and 2090 using the mean and standard deviation of Z:
Standardized value of 1970 = (1970 - E(Z)) / (Standard deviation of Z) = (1970 - 2000) / 50.99 ≈ -0.59
Standardized value of 2090 = (2090 - E(Z)) / (Standard deviation of Z) = (2090 - 2000) / 50.99 ≈ 1.76
Next, we need to calculate the probability of the standardized values falling within this range using a standard normal distribution table or calculator:
P(-0.59 < Z < 1.76)
Looking up the values in a standard normal distribution table, we find the cumulative probabilities corresponding to the standardized values:
P(Z < -0.59) ≈ 0.2776
P(Z < 1.76) ≈ 0.9599
To find the probability of the range, we subtract the lower probability from the higher probability:
P(-0.59 < Z < 1.76) ≈ P(Z < 1.76) - P(Z < -0.59) ≈ 0.9599 - 0.2776 ≈ 0.6823
Therefore, the approximate probability for 1970 < Z < 2090 is approximately 0.6823.