On a normally distributed anxiety test with mean 48 and standard deviation 4, approximately what anxiety test score would put someone in the top 5 percent?

54.56

Step-by-step explanation:

Given:

Mean, u = 48

Standard deviation, = 4

Test score = 5%

Required:

Find the raw score, X.

Having a test score of 5% means the Zscore has to correspond to 95%, ie, 1 - 0.05 = 0.95.

Thus,

To get X, use the formula:

= 48 + (4 * 1.64)

= 48 + 6.56

= 54.56

The anxiety test score that would put someone in the top 5% is 54.56

linearly: 8560 bacteria

exponentially: approximately 75233 bacteria

step-by-step explanation:

assuming it is not exponential (we would have some really complicated logarithms), let's say it is linear. then, after 5 hours, it increased by 1200 bacteria, so every hour it increases by 240 bacteria. this means that after 19 hours, we get 19 * 240 + 4000 bacteria or 3800 + 760 + 4000 = 7800 + 760 = 8560 bacteria.

because it says round your answer, i will do this exponentially (with a calculator, of course). first, we need to find how much it multiplies by every hour, by taking the 5th root of 52/40 or 13/10. this is approximately 0.167. we still need to add 1 to this, because we are multiplying by that number, not adding that amount of the number. now, we just calculate 4000 * (1.167)^19. use a calculator to get approximately 75232.6, or 75233.

560

step-by-step explanation:

giggidy