Answer:
The probability is 0.0014
Step-by-step explanation:
When we have two possible outcomes p, q and we repeat the experiment multiple times, we apply the Binomial distribution, in this distribution the probability of getting exactly k successes in n trials is given by
where p represents the probability of success. nCk refers to the combinations of k elements out of n elements.
In this case, p= .29 (the probability that the person says "yes" being confident that passenger trips to the moon will occur in their lifetime.
q = .71 (The probability that the person says "no")
We sample 200 americans, therefore, we have n = 200 and we need the probability that less than 40% of the people sampled answer "yes".
The 40% of 200 is 80. So, in other words, we need less than 80 people to answer yes to the question.
P(x<80) = P(x=0) + P(x=1) + P(x=2) + .... + P(x=79) = 0.00143466294 (we can use a Binomial Distribution calculator to compute this result)
Therefore, P(x < 80) = 0.0014
Answer:
Step-by-step explanation:
Given : A biologist studied how cells grow by adding liquid protein to cell samples.
She divided of an ounce of liquid protein evenly among 3 cell samples.
Here , total quantity of liquid protein = of an ounce
Total cell samples = 3
Now , Quantity of liquid protein given by her to each cell = (total quantity of liquid protein ) ÷ (3)
= .
Hence, Quantity of liquid protein given by her to each cell =
Answer:
a) 0.7287
b) 0.9663
c) 0.237
d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4.5 million tons of cargo per week
Standard Deviation, σ = 0 .82 million tons
We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.
Formula:
a) P( port handles less than 5 million tons of cargo per week)
P(x < 5)
Calculation the value from standard normal z table, we have,
b) P( port handles 3 or more million tons of cargo per week)
Calculating the value from the standard normal table we have,
c)P( port handles between 3 million and 4 million tons of cargo per week)
d) P(X=x) = 0.85
We have to find the value of x such that the probability is 0.85.
P(X > x)
Calculation the value from standard normal z table, we have,
Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.