Suppose the mean income of firms in the industry for a year is 90 million dollars with a standard deviation of 7 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 97 million dollars? Round your answer to four decimal places.
I need help learning how to do this on a ti-84 plus calculator. Could someone please help out?

For the rule of 78, for a 12-month period, the last term in the sequence is 12 and the series sums to 78. for an 10 month period, the last term is and the series sum is . for a 15 month period, the last term is and the series sum is . for a 20 month period, the last term is and the series sum is

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Each of the following data sets has a mean of x = 10. (i) 8 9 10 11 12 (ii) 7 9 10 11 13 (iii) 7 8 10 12 13 (a) without doing any computations, order the data sets according to increasing value of standard deviations. (i), (iii), (ii) (ii), (i), (iii) (iii), (i), (ii) (iii), (ii), (i) (i), (ii), (iii) (ii), (iii), (i) (b) why do you expect the difference in standard deviations between data sets (i) and (ii) to be greater than the difference in standard deviations between data sets (ii) and (iii)? hint: consider how much the data in the respective sets differ from the mean. the data change between data sets (i) and (ii) increased the squared difference îł(x - x)2 by more than data sets (ii) and (iii). the data change between data sets (ii) and (iii) increased the squared difference îł(x - x)2 by more than data sets (i) and (ii). the data change between data sets (i) and (ii) decreased the squared difference îł(x - x)2 by more than data sets (ii) and (iii). none of the above