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

43

step-by-step explanation:

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)