Suppose x has a distribution with = 10 and = 2. (a) If a random sample of size n = 46 is drawn, find x, x and P(10 ≤ x ≤ 12). (Round x to two decimal places and the probability to four decimal places.)

x = ?
x = ?
P(10 ≤ x ≤ 12) = ?

(b) If a random sample of size n = 73 is drawn, find x, x and P(10 ≤ x ≤ 12). (Round x to two decimal places and the probability to four decimal places.)
x = ?
x = ?
P(10 ≤ x ≤ 12) = ?

(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is part (a) because of the sample size. Therefore, the distribution about x is .

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

stops on an even number.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

if you would like with a question it is easy to get an answer by including some or multi choice options

step-by-step explanation: