Suppose a binomial random variable consists of 10 trials with a success of probability 0.35. what are the mean and standard division for this binomial random variable?
the mean is 3.5 and the deviation is 2.275
the mean is 3.5 and a standard deviation is 1.508.
the mean is 10 and the standard deviation is 0.35.
the mean is 0 and the standard deviation is 1
none of the other answers are correct

For 96% confidence interval for the population mean of miles driven :

Lower bound  = 10,841 miles

Upper bound= 14,949 miles

Step-by-step explanation:

Here, population standard deviation is unknown and sample size is small , So the formula is used to find the confidence interval for is given by :-

, where n = sample size , = sample mean , t*= two tailed critical value s= sample standard deviation, .

Given, =12,895 miles , s=3,801 miles, n=15 , degree of freedom = 14 [∵df=n-1]

For 96% confidence level ,

By t-distribution table ,

t-value(two tailed) for and df =14 is t*=2.2638

Now , the 96% confidence interval for the population mean of miles driven will be :

Hence, For 96% confidence interval for the population mean of miles driven :

lower bound  = 10,841 miles

upper bound= 14,949 miles