Answer:
1. 80
2. 80
3. 40
Step-by-step explanation:
The question requires us to Use the sample size of 200 to find the expected frequency for categories A, B and C
1. (Category proportion)(sample size)
eA = 0.40x200
= 80
2. (Category proportion)(sample size)
eB = 0.40x200
= 80
3. (Category proportion)(sample size)
eC = 0.20 x 200
= 40
The answers are 80, 80 and 40 respectively.
Answer:
A) probability of failure in next 100 hours given that it has been tested for 500 hours without failure is 0.181
B) probability that exactly two have the metabolic defect is 0.03
Step-by-step explanation:
Part A)
Let X be a exponentially random variable with mean = μ = 500 hrs
For exponential distribution:

λ = 1/μ
λ = 0.002
We have to find the probability of failure in the next 100 hours given that assembly has been tested for 500 hours without a failure.
Using memory less property of exponential distribution:

using

<h3>Part B)</h3>
Chances of occurrence of metabolic defect = 5%
P(C) = .05
No. of randomly selected infants = n =6
We have to find the probability that exactly two have the metabolic defect
⇒x = 2
Using binomial probability density function:
P = ![P=\left[\begin{array}{ccc}n\\x\end{array}\right] p^{x} (1-p) ^{n-x}\\\\=\frac{n!}{x!(n-x)!} p^{x} (1-p) ^{n-x}\\=\frac{6!}{2!4!}(.05)^{2}(.95)^{4}\\= 0.03\\](https://tex.z-dn.net/?f=P%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dn%5C%5Cx%5Cend%7Barray%7D%5Cright%5D%20p%5E%7Bx%7D%20%281-p%29%20%5E%7Bn-x%7D%5C%5C%5C%5C%3D%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D%20p%5E%7Bx%7D%20%281-p%29%20%5E%7Bn-x%7D%5C%5C%3D%5Cfrac%7B6%21%7D%7B2%214%21%7D%28.05%29%5E%7B2%7D%28.95%29%5E%7B4%7D%5C%5C%3D%200.03%5C%5C)
probability that exactly two have the metabolic defect is 0.03
Answer:
imm a bit sorry I could only answer the international one as the national is equal to international place value system
Step-by-step explanation:
eight hundred ninety three thousand four hundred fifty one
hth tth th h t o
8 9 3 4 5 1
I hope you understand