Answer:
probability that all of the sprinklers will operate correctly in a fire: 0.0282
Step-by-step explanation:
In order to solve this question we will use Binomial probability distribution because:
- In the question it is given that the sprinklers activate correctly or not independently.
- The number of outcomes are two i.e. sprinklers activate correctly or not.
A binomial distribution is a probability of a success or failures outcomes in an repeated multiple or n times.
Number of outcomes of this distributions are two.
The formula is:
b(x; n, P) = 
b = binomial probability also represented as P(X=x)
x =no of successes
P = probability of a success on a single trial
n = no of trials
is calculated as:
= n! / x!(n – x)!
= 10! / 10!(10-10)!
= 1
According to given question:
probability of success i.e. p = 0.7 i.e. probability of a sprinkler to activate correctly.
number of trials i.e. n = 10 as number of sprinklers are 10
To find: probability that all of the sprinklers will operate correctly in a fire
X = 10 because we have to find the probability that "all" of the sprinklers will operate correctly and there are 10 sprinklers so all 10 of them
So putting these into the formula:
P(X=x) =
= C₁₀,₁₀ * 0.7¹⁰ * (1-0.7)¹⁰⁻¹⁰
= 1 * 0.0282 * (0.3) ⁰
= 1 * 0.0282 * 1
P(X=x) = 0.0282
Answer:
a)0.099834
b) 0
Step-by-step explanation:
To solve for this question we would be using , z.score formula.
The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.
A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.
a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.
Standard Deviation = √variance
= √0.16 = 0.4
Standard deviation = 0.4
Mean = 21.37
x = 20.857
z = (x-μ)/σ
z = 20.857 - 21.37/0.4
z = -1.2825
P-value from Z-Table:
P(x<20.857) = 0.099834
b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.
z score formula used = (x-μ)/σ/√n
x = 20.857
Standard deviation = 0.4
Mean = 21.37
n = 100
z = 20.857 - 21.37/0.4/√100
= 20.857 - 21.37/ 0.4/10
= 20.857 - 21.37/ 0.04
= -12.825
P-value from Z-Table:
P(x<20.857) = 0
c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.
The answer is D. <span>No, it is not a valid inference because his classmates do not make up a random sample of the students in the school.</span>
Answer:
Step-by-step explanation:
The side of the square = 3^2/7
Use the law of exponents . If the power is a fraction, that means it is
3^2/7 = 3^2 x 1/7 = 7√9
To find the area you multiply this by itself.
This gives you 1.87...
Hope this helps
