Aset of exam scores is normally distributed and has a mean of 74.4 and a standard deviation of 8.5. what is the probability that a randomly selected score will be between 69 and 77?

answer = (round to four decimal places)

note: be use the z table not use technology or the 68-95-99.7 rule.

Step-by-step explanation:

Previous concepts  

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".  

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".    

Let X the random variable who represent the exam scores for the population, and for this case we know the distribution for X is given by:  

Where and  

And let represent the sample mean, the distribution for the sample mean is given by:  

What is the probability that a randomly selected score will be between 69 and 77?

For this case we can use the z score formula given by:

And that correspond with the 35.75% of the data.

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.

answer: haha (: i’ll go see what it is

step-by-step explanation: