I dont know my math anymore
2 answers:
You might be interested in
Answer:
A) 0.5737
B) 0.9884
Step-by-step explanation:
We are given that an engineer is going to redesign an ejection seat for an airplane. The new population of pilots has normally distributed weights with a mean of 160 lb and a standard deviation of 27.5 lb i.e.; = 160 lb and
= 27.5 lb
(A) We know that Z = ~ N(0,1)
Let X = randomly selected pilot
If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)
P(150 < X < 201) = P(X < 201) - P(X <= 150)
P(X < 201) = P( <
) = P(Z < 1.49) = 0.9319
P(X <= 150) = P( <
) = P(Z < -0.3636) = P(Z > 0.3636) = 0.3582
Therefore, P(150 < X < 201) = 0.9319 - 0.3582 = 0.5737 .
(B) We know that for sampling mean distribution;
Z = ~ N(0,1)
If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);
P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)
P(X bar < 201) = P( <
) = P(Z < 9.311) = 1 - P(Z >= 9.311)
= 0.999995
P(X bar <= 150) = P( <
) = P(Z < -2.2709) = P(Z > 2.2709)
= 0.0116
Therefore, P(150 < X bar < 210) = 0.999995 - 0.0116 = 0.9884
C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.
<span>For the plane, we have z = 5x + 9y
For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.
Since x > x^4 for y in [0, 1],
The volume of the solid equals
![\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx \\ \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx \\ \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\ \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E1_0%20%7B%20%5Cint%5Climits_%7Bx%5E4%7D%5Ex%20%7B%285x%2B9y%29%7D%20%5C%2C%20dy%20%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B5xy%2B%20%5Cfrac%7B9%7D%7B2%7D%20y%5E2%5Cright%5D_%7Bx%5E4%7D%5E%7Bx%7D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%5B%5Cleft%285x%28x%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%29%5E2%5Cright%29-%5Cleft%285x%28x%5E4%29%2B%20%5Cfrac%7B9%7D%7B2%7D%20%28x%5E4%29%5E2%5Cright%29%5Cright%5D%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%285x%5E2%2B%20%5Cfrac%7B9%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E1_0%20%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D%20x%5E2-5x%5E5-%20%5Cfrac%7B9%7D%7B2%7D%20x%5E8%5Cright%29%7D%20%5C%2C%20dx%20%5C%5C%20%20%5C%5C%20%3D%5Cleft%5B%20%5Cfrac%7B19%7D%7B6%7D%20x%5E3-%20%5Cfrac%7B5%7D%7B6%7D%20x%5E6-%20%5Cfrac%7B1%7D%7B2%7D%20x%5E9%5Cright%5D%5E1_0)
</span>
For the region, we first find its boundary curves' points of intersection.
x = x^4 ==> x = 0, 1.
Since x > x^4 for y in [0, 1],
The volume of the solid equals