The amount of time spent by North American adults watching television per day is normally distributed with a mean of 6 hours and a standard deviation of 1.5 hours.
What is the probability that in a random sample of five North American adult at least two of them watch television for more than 7 hours per day?
To find the probability that at least two North American adults watch television for more than 7 hours per day in a random sample of five, we need to calculate the probability of the complementary event, which is the probability that less than two adults watch television for more than 7 hours per day.
First, let's find the probability that one North American adult watches television for more than 7 hours per day. We can use the standard normal distribution to calculate this probability.
The z-score formula is:
z = (x - μ) / σ
Where:
x = the value we want to find the probability for (7 hours)
μ = the mean (6 hours)
σ = the standard deviation (1.5 hours)
Substituting the values:
z = (7 - 6) / 1.5
z = 1 / 1.5
z ≈ 0.67
Using a standard normal distribution table (z-table) or a calculator, we find that the probability corresponding to a z-score of 0.67 is approximately 0.7486.
Next, we need to calculate the probability that less than two adults (0 or 1) watch television for more than 7 hours per day. To do this, we can use the binomial distribution formula:
P(X < 2) = P(X = 0) + P(X = 1)
Where:
P(X < 2) = probability that less than two adults watch television for more than 7 hours per day
P(X = 0) = probability that no adults watch television for more than 7 hours per day
P(X = 1) = probability that exactly one adult watches television for more than 7 hours per day
The binomial distribution formula is:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
C(n, k) = the number of combinations of n items taken k at a time
n = the total number of trials (sample size) = 5 in this case
k = the number of successes (adults watching television for more than 7 hours per day) = 0 or 1
p = the probability of a success (probability of an adult watching television for more than 7 hours per day) = 0.7486
(1-p) = the probability of a failure (probability of an adult watching television for less than or equal to 7 hours per day) = 1 - 0.7486 = 0.2514
Calculating the probability for each case:
P(X = 0) = C(5, 0) * 0.7486^0 * 0.2514^5 = 1 * 1 * 0.2514^5 = 0.0025
P(X = 1) = C(5, 1) * 0.7486^1 * 0.2514^4 = 5 * 0.7486^1 * 0.2514^4 ≈ 0.0317
Therefore,
P(X < 2) = P(X = 0) + P(X = 1) = 0.0025 + 0.0317 ≈ 0.0342
Now, we can find the probability that at least two adults watch television for more than 7 hours per day by subtracting this probability from 1:
P(at least two adults watch > 7 hours) = 1 - P(X < 2) = 1 - 0.0342 ≈ 0.9658
Hence, the probability that in a random sample of five North American adults at least two of them watch television for more than 7 hours per day is approximately 0.9658 or 96.58%.