A survey of several 1010 to 1111 year olds recorded the following amounts spent on a trip to the mall: $18.31,$25.09,$26.96,$26.54,$21.84, $21.46$ 18.31,$ 25.09,$ 26.96,$ 26.54,$ 21.84,$ 21.46 Construct the 98%98% confidence interval for the average amount spent by 1010 to 1111 year olds on a trip to the mall. Assume the population is approximately normal.

Step-by-step explanation:

Hello!

The variable of interest is

X: the amount a 10-11-year-old spends on a trip to the mall.

Assuming that the variable has a normal distribution, you have to construct a 98% CI for the average of the amount spent by the 10-11 year-olds on one trip to the mall.

For this you have to use a Student-t for one sample:

X[bar] ± *

n= 6

$18.31, $25.09, $26.96, $26.54, $21.84, $21.46

∑X= 140.20

∑X²= 3333.49

X[bar]= ∑X/n= 140.20/6= 23.37

S²= 1/(n-1)*[∑X²-(∑X)²/n]= 1/5[3333.49-(140.2)²/6]= 11.50

S= 3.39

X[bar] ± *

[23.37 ± 3.365 * ]

[18.66;29.98]

With a confidence level of 98%, you'd expect that the interval $[18.66;29.98] will include the population mean of the money spent by 10-11 year-olds in one trip to the mall.

I hope you have a SUPER day!

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

it is x=-4

step-by-step explanation: