Jim tree sells trees. the mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. jim wants to know what percent of his sales were more than 84 inches tall. he can use the standard normal distribution to him. he asks, "what is the mean? " mean = inches. he thinks, "what is the expected distribution about the mean? " ( 38 inches 48 inches 58 inches 68 inches 78 inches 88 inches 98 inches. ) he thinks, " % of the sales will be below 68 inches." but, he needs to know the percentage between 68 and 84 to add to that. so, he calculates a z-score = = , and finds the percentage associated with 1.6 in the table. this is (to the nearest tenth) %. now jim knows that % of his sales were 84 inches or less. therefore, the remaining % were more than 84 inches.

Jim Tree sells trees. The mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. Jim wants to know what percent of his sales were more than 84 inches tall. He can use the standard normal distribution to help him.

He asks, "what is the mean?" Mean = 68 inches.

He thinks, "What is the expected distribution about the mean?" ( 38 inches 48 inches 58 inches 68 inches 78 inches 88 inches 98 inches.)

He thinks, " 50% of the sales will be below 68 inches."

But, he needs to know the percentage between 68 and 84 to add to that. So, he calculates a z-score
z-score=(x-mean)/(standard deviation)=(84-68)/10=16/10
z-score=1.6

and finds the percentage associated with 1.6 in the table.
0.35543=0.35543*100%=35.543%=35.5%

 This is (to the nearest tenth) 35.5%.

Now Jim knows that 50%+35.5%=85.5% of his sales were 84 inches or less.

Therefore, the remaining 100%-85.5%=14.5% were more than 84 inches.