Given:
μ = 68 in, population mean
σ = 3 in, population standard deviation
Calculate z-scores for the following random variable and determine their probabilities from standard tables.
x = 72 in:
z = (x-μ)/σ = (72-68)/3 = 1.333
P(x) = 0.9088
x = 64 in:
z = (64 -38)/3 = -1.333
P(x) = 0.0912
x = 65 in
z = (65 - 68)/3 = -1
P(x) = 0.1587
x = 71:
z = (71-68)/3 = 1
P(x) = 0.8413
Part (a)
For x > 72 in, obtain
300 - 300*0.9088 = 27.36
Answer: 27
Part (b)
For x ≤ 64 in, obtain
300*0.0912 = 27.36
Answer: 27
Part (c)
For 65 ≤ x ≤ 71, obtain
300*(0.8413 - 0.1587) = 204.78
Answer: 204
Part (d)
For x = 68 in, obtain
z = 0
P(x) = 0.5
The number of students is
300*0.5 = 150
Answer: 150
Answer:
Cov(X, Y) =0.029.
Step-by-step explanation:
Given that :
The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ............(1)
Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.
0.04e–jτj/10 ............(2)
Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.
That is, they are 5 seconds apart,
τ = 5 seconds..............(3)
Thus,
Cov(X, Y), for τ = 5seconds = 0.04e-5/10
= 0.04e-0.5 = 0.04/√e
= 0.04/1.6487
= 0.0292
Thus, Cov(X, Y) =0.029.
The volume of a cube is found by multiplying the length of any edge by itself twice. So if the length of an edge is 4, <span>the volume is 4 x 4 x 4 = 64</span>
Answer:
Pr(X-Y ≤ 44.2) = 0.5593
Step-by-step explanation:
for a certain breed of terrier
Mean(μ) = 72cm
Standard deviation (σ) = 10cm
n = 64
For a certain breed of poodle
Mean(μ) = 28cm
Standard deviation (σ) = 5cm
n = 100
Let X be the random variable for the height of a certain breed of terrier
Let Y be the random variable for the height of a certain breed of poodle
μx - μy = 72 -28
= 44
σx - σy = √(σx^2/nx + σy^2/ny)
= √10^2/64 + 5^2/100
= √100/64 + 25/100
= √ 1.8125
= 1.346
Using normal distribution,
Z= (X-Y- μx-y) / σx-y
Z= (44.2 - 44) / 1.346
Z= 0.2/1.346
Z= 0.1486
From the Z table, Z = 0.149 = 0.0593
Φ(z) = 0 0593
The probability that the difference of the observed sample mean is at most 44.3 is Pr(Z ≤ 44.2)
Recall that if Z is positive,
Pr(Z≤a) = 0.5 + Φ(z)
Pr(Z ≤ 44.2) = 0.5 + 0.0593
= 0.5593
Therefore,
Pr(X-Y ≤ 44.2) = 0.5593
