Question:
Consider the following exponential probability density function.
f(x) = 1/5e^(−x/5) for x ≥ 0
(a) Write the formula for P(x ≤ x0). (b) Find P(x ≤ 4). (Round your answer to four decimal places.) (c) Find P(x ≥ 5). (Round your answer to four decimal places.) (d) Find P(x ≤ 6). (Round your answer to four decimal places.) (e) Find P(4 ≤ x ≤ 6). (Round your answer to four decimal places.)
Answer:
(a) P(x ≤ x0) = 1 - e^(−x0/5)
(b) P(x ≤ 4) = 0.5506
(c) P(x ≥ 5) = 0.3678
(d) P(x ≤ 6) = 0.6988
(e) P(4 ≤ x ≤ 6) = 0.1482
Step-by-step explanation:
The standard form of the exponential probability density function is given by
f(x) = 1/μe^(−x/μ)
Where μ is the mean, for the given problem μ = 5
(a) Write the formula for P(x ≤ x0)
P(x ≤ x0) = 1 - e^(−x0/5)
(b) Find P(x ≤ 4)
P(x ≤ 4) = 1 - e^(−4/5)
P(x ≤ 4) = 1 - 0.4493
P(x ≤ 4) = 0.5506
(c) Find P(x ≥ 5)
P(x ≥ 5) = e^(−5/5)
P(x ≥ 5) = 0.3678
(d) Find P(x ≤ 6)
P(x ≤ 6) = 1 - e^(−6/5)
P(x ≤ 6) = 1 - 0.3011
P(x ≤ 6) = 0.6988
(e) Find P(4 ≤ x ≤ 6)
P(4 ≤ x ≤ 6) = e^(−4/5) - e^(−6/5)
P(4 ≤ x ≤ 6) = 0.4493 - 0.3011
P(4 ≤ x ≤ 6) = 0.1482
Answer:
20 minutes
Step-by-step explanation:
Melissa works as fast as two Bettys, so working together, they get the job done at the rate 3 Bettys could do it. That time is (60 minutes)/3 = 20 minutes.
Working together, Betty and Melissa can mow the lawn in 20 minutes.
_____
You can also think in terms of mowing rates as lawns per minute. The total mowing rate is ...
Betty's rate + Melissa's rate = (1/60 lawns/min) +(1/30 lawns/min)
= (1/60 +2/60) lawns/min = 1/20 lawns/min
The inverse rate is then ...
20/1 min/lawn
Together, they take 20 minutes to mow 1 lawn.
Answer:
I believe the answer would be using the form of y= mx+b, b is y intercept and slope is m. It would end up looking like y= 2/3x+1
Step-by-step explanation:
I think this is how a linear equation looks like but I'm a little rusty. If you have notes it probably says what I already stated. Good luck! :)
Dang...H0: μ = 115
HA: μ ≠ 115
Sample mean = 120
Standard deviation = 25
Standard error of mean = σ / √ n
Standard error of mean = 25 / √ 100
SE = 25/10
Standard error of mean 2.5
z = (xbar- μ ) / SE
z = (120-115) / 2.5
z = 2
p-value = 2 P( z > 2) = 2(0.0228) = 0.0456
the data are statistically significant at level = .05, but not at level = .01.
2)
H0: μ = 115
HA: μ ≠ 115
Sample mean = 119
Standard deviation = 25
Standard error of mean = σ / √ n
Standard error of mean = 25 / √ 100
SE = 25/10
Standard error of mean 2.5
z = (xbar- μ ) / SE
z = (119-115) / 2.5
z = 1.6
p-value = 2P( z > 1.6) = 2(0.0548) =0.1096
3)
a statement about the population the researcher suspects is true, and is trying to find evidence for.
4)
Sample mean = 80
Standard deviation = 20
Standard error of mean = σ / √ n
Standard error of mean = 20 / √ 100
SE = 20/10
The Standard error of mean 2
Confidence interval 80-(2)(1.645)
and 80+(2)(1.645)
(76.7, 83.3)
