Answer:
the means absolute deviation is 1.8
the striking deviation is 7
(it is because the data point on the far right of the graph)
Step-by-step explanation:
calculate the absolute deviation:
1. calculate the mean (add all numbers and divide by 10) = 2.6
2. find the absolute value of each
2.6 - 0 = 2.6
2.6 - 1 = 1.6 (you have 1's 4 times so each one will be 1.6)
2.6 - 3 = 0.4
2.6 - 4 = 1.4 (you have 4's 3 times)
2.6 - 7 = 4.4
3. Then add all of those values together
2.6 + 1.6 + 1.6 + 1.6 + 1.6 + 0.4 + 1.4 + 1.4 + 1.4 + 7.4 = 18
4. find the mean of the difference
18/10 = 1.8
Sold 2/5 in the morning.....leaving 3/5
he sold 3/4 of 3/5 in the afternoon......3/4 * 3/5 = 9/20
2/5 = 8/20
9/20 - 8/20 = 1/20...so he sold 1/20 more in the afternoon.
so if 1/20 = 24, then 20/20 = 24*20 = 480
so he started with 480 cartons
1. The fourth graph: D
2. The first graph: A
Answer:
3/4 or 75% are boys
Step-by-step explanation:
Answer:
a) For the 90% confidence interval the value of 
and 
, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
b) For the 99% confidence interval the value of 
and 
, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:
Step-by-step explanation:
Previous concepts
The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."
Solution to the problem
Part a
For the 90% confidence interval the value of and , with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:
Part b
For the 99% confidence interval the value of and , with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:
