Answer:
Step-by-step explanation:
Hello!
To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.
To test this claim two samples were taken:
Sample 1
X: battery lifespan of a unit of the new product (min)
n= 93 units of the new product
mean battery life X[bar]= 8:53hs= 533min
S= 84 min
Sample 2
X: battery lifespan of a unit of the leading product (min)
n= 102 units of the leading product
mean battery life X[bar]= 5:40 hs = 340min
S= 93 min
The population variances of both variances are unknown and distinct.
To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:
H₀: μ₁ - μ₂ ≤ 120
H₁: μ₁ - μ₂ > 120
Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:
![Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }](https://tex.z-dn.net/?f=Z%3D%20%5Cfrac%7B%28X%5Bbar%5D_1-X%5Bbar%5D_2%29-%28Mu_1-Mu_2%29%7D%7B%5Csqrt%7B%5Cfrac%7BS_1%5E2%7D%7Bn_1%7D%20%2B%5Cfrac%7BS_2%5E2%7D%7Bn_2%7D%20%20%7D%20%7D)
Z≈N(0;1)
I hope this helps!
Answer:
The answers to the questions are;
(a) P(At least 1 defective)
= 0.9883.
(b) P(At least 1 defective)
= 0.6409.
Step-by-step explanation:
There are 110 cards and 20 defectives.
a) The probability of at least one defective is given by
P(At least 1 defective) = 1 - P(0 defective)
P(0 defective) = 20C0 × (90C0)/(110C20) = 0.0116
1 - 0.0116 = 0.9883
b) For a set of 110 boards that has 5 defective and 105 non-defective
P(At least 1 defective) = 1 - P(0 defective)
P(0 defective) = (20C0)(90C5)/(110C5) = 0.35909
1-0.35909
= 0.6409
Answer:
0.9355
Step-by-step explanation:
What we will use here is conditional probability formula.
let A be the event that the plane departs on time
and B be the event that it arrives on time
P(A) = 0.87
P(B|A) = 0.93
P(B) = ?
P(A n B) = ?
Mathematically;
P(B|A) = P(B nA)/P(A)
0.93 = 0.87/P(A)
P(A) = 0.87/0.93
P(A) = 0.935483870967742
which is 0.9355 to four decimal places
Answer:
1: 119 2: 61 3: 61 4: 119 5: 119 6: 61 7: 61
Step-by-step explanation:
Since we know that m & n are parallel we can prove which angles are congruent to angle 8 and which are supplementary
5: 119 (vertical to angle 8)
1: 119 (corresponding angles theorem
2: 61 (linear pair)
3: 61 (vertical to 2)
4: 119 (vertical to 1)
6: 61 (same sided interior angles theorem)
7: 61 (corresponding)
Answer 1: 
Answer 2: 
Note: BOTH of those answers are correct. Chose either one.
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<u>Set factors equal to 0</u>
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