Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 11 points and an unknown population mean. A random sample of 15 scores is taken and gives a sample mean of 101 points. Find the confidence interval for the population mean with a 98% confidence level.

So on this case the 98% confidence interval would be given by (94.382;107.618)  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

represent the sample mean  

population mean (variable of interest)  

represent the population standard deviation  

n=15 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

(1)  

The point estimate of the population mean is

Since the Confidence is 0.98 or 98%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that  

Now we have everything in order to replace into formula (1):  

So on this case the 98% confidence interval would be given by (94.382;107.618)  

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

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.