1.A magazine reported that 6% of Americans read the news paper while driving. 100 drivers are selected at random, find the probability that exactly 25 of them say they read the newspaper while driving.(normal binomial formula to solve)
2. A survey of 30 adults found that the mean age of a person's primary vehicle is 5.6 years. Assuming that the standard deviation of the population is 0.8 years find best point estimate of the population is 0.8 years find the best point estimate of the population mean and the 99% confidence interval of that population mean.
3.ten randomly selected people were asked how long they slept at night. The mean time was 7.1 hours and the (sample) standard deviation was 0.78 hours. Find the 95% confidence interval of the mean time. Assume the variable is normally distributed.
1. To find the probability that exactly 25 out of 100 drivers say they read the newspaper while driving, we can use the normal binomial distribution formula.
The formula for the probability mass function (PMF) of a binomial distribution is:
P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n is the total number of trials or drivers selected (in this case, n = 100)
- k is the number of successful outcomes or drivers who read the newspaper while driving (k = 25)
- p is the probability of success for each trial (p = 0.06, as stated in the problem)
Using this formula, we can calculate the probability as follows:
P(X=25) = (100 choose 25) * (0.06^25) * (1-0.06)^(100-25)
To calculate this, you would need to use a combination calculator to find the value of (100 choose 25), and then substitute the values into the formula.
2. To find the best point estimate of the population mean and the 99% confidence interval, we can use the sample mean and the standard deviation.
The best point estimate of the population mean can be calculated simply by using the sample mean. In this case, the sample mean is 5.6 years.
To calculate the confidence interval, we can use the formula:
Confidence Interval = sample mean +/- (critical value * (standard deviation / sqrt(n)))
In this case, since the population standard deviation is known to be 0.8 years, we can use the z-score to find the critical value for a 99% confidence level. The z-score corresponding to a 99% confidence level is approximately 2.576.
Substituting the values into the formula, we get:
Confidence Interval = 5.6 +/- (2.576 * (0.8 / sqrt(30)))
Calculating this will give you the range within which the true population mean is estimated to lie with 99% confidence.
3. To find the 95% confidence interval of the mean sleep time, we can use the sample mean and the sample standard deviation.
The formula for the confidence interval is similar to the one mentioned in Question 2:
Confidence Interval = sample mean +/- (critical value * (sample standard deviation / sqrt(n)))
In this case, we use the sample mean of 7.1 hours and the sample standard deviation of 0.78 hours. Since the variable is assumed to be normally distributed, we can use the t-distribution to find the critical value for a 95% confidence level with nine degrees of freedom (n-1).
Using a t-table or calculator, we can find the t-value associated with a 95% confidence level and nine degrees of freedom, which is approximately 2.262.
Substituting the values into the formula, we get:
Confidence Interval = 7.1 +/- (2.262 * (0.78 / sqrt(10)))
Calculating this will give you the range within which the true population mean sleep time is estimated to lie with 95% confidence.
1. To find the probability that exactly 25 out of 100 randomly selected drivers say they read the newspaper while driving, we can use the binomial probability formula.
The formula is: P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- P(X=k) is the probability of getting exactly k successes
- n is the number of trials or sample size
- p is the probability of success in each trial
In this case, n = 100, k = 25, and p = 0.06 (6%). Substituting these values into the formula, we get:
P(X=25) = (100 choose 25) * (0.06)^25 * (1-0.06)^(100-25)
= (100! / (25!(100-25)!)) * (0.06^25) * (0.94^75)
Using a calculator or statistical software, you can calculate this probability.
2. To find the best point estimate of the population mean and the 99% confidence interval, we can use the sample mean and the formula for confidence intervals.
The best point estimate of the population mean can be obtained using the sample mean. In this case, the sample mean is 5.6 years.
For the confidence interval, the formula is:
CI = sample mean +/- (critical value * standard deviation / square root of sample size)
To find the critical value for a 99% confidence interval, we need to look it up in a standard normal distribution table or use software. The critical value for a 99% confidence interval is approximately 2.576.
The standard deviation given is 0.8 years, and the sample size is 30.
Plugging these values into the formula, we have:
CI = 5.6 +/- (2.576 * 0.8 / sqrt(30))
Calculate the above expression to get the confidence interval.
3. To find the 95% confidence interval of the mean time slept at night, we can use the sample mean, sample standard deviation, and the formula for confidence intervals.
The formula is:
CI = sample mean +/- (critical value * sample standard deviation / square root of sample size)
The sample mean given is 7.1 hours, the sample standard deviation is 0.78 hours, and the sample size is 10.
To find the critical value for a 95% confidence interval, we need to look it up in a t-distribution table or use software. For simplicity, let's assume the sample size is large enough to use a z-distribution. The critical value for a 95% confidence interval is approximately 1.96.
Plugging these values into the formula, we have:
CI = 7.1 +/- (1.96 * 0.78 / sqrt(10))
Calculate the above expression to get the confidence interval.