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. 1. 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. ) 2. 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. 3. this is (to the nearest %. now jim knows that % of his sales were 84 inches or less. 4. therefore, the remaining % were more than 84 inches.

1. The mean is given in the problem statement as 68 inches.

2. 50% of the normal distribution is below the mean. This question answers itself regarding the z-score. It is 1.6. (Jim computed it as z = (x-μ)/σ = (84-68)/10 = 1.6.)

3. We don't have your table, so we don't know what it tells you. We expect it will tell you either 0.05480 or 0.94520. Assuming the latter, the percentage associated with 1.6 in the table is 94.5%. Now Jim knows 94.5% of his sales were 84 inches or less.

4. Therefore, the remaining 5.5% were more than 84 inches.

The numbers more than and less than a given value must total 100%.

step-by-step explanation:

this woman was an important writer and propagandist during the american revolution. she was a member of the committee of correspondence (in fact the committee was formed in her home) and was an advisor to many of the revolutionary leaders, including george washington and john adams. who was this revolutionary woman?

no this is a nonlinear negative