Suppose the salaries of university professors are approximately normally distributed with a mean of $65,000 and a standard deviation of $7,000. If a random sample of size 25 is taken and the mean is calculated, what is the probability that the mean value will be between $62,500 and $64,000? a. .1465 b. .0827 c. .0371 d. .2005

0.2005

Step-by-step explanation:

Mean, m = 65000

Standard deviation, σ= 7000

Sample size, n = 25

Let X = random variable of salary

Recall:

Z = (μ - x) /(σ/√n)

P(62500 ≤ x ≤ 64000) =?

Pr((65000 - 62500)/7000/√25 ≤ z ≤ (65000 - 64000) / 7000/√25)

P(2500 / 1400 ≤ z ≤ 1000/1400)

P = (1.79 ≤ z ≤ 0.714)

Using the normal distribution table or a Z probability calculator

0.4633 - 0.2624

= 0.2009

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.

answer: 3.26019516

step-by-step explanation: