Answer:
Correct graph
Incorrect graph
Slopes are not the same
Ms. JEFFERSON SPENT 16 DOLLARS TO BUY 12.8 OUNCES OF SMOKED TROUT. WHAT WAS THE COST PER Ounce?
1 answer:
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Answer:
81.86%
Step-by-step explanation:
We have been given that final exam scores are normally distributed with a mean of 74 and a standard deviation of 6.
First of all we will find z-score using z-score formula.
Now let us find z-score for 86.
Now we will find P(-1<Z) which is probability that a random score would be greater than 68. We will find P(2>Z) which is probability that a random score would be less than 86.
Using normal distribution table we will get,
We will use formula to find the probability to find that a normal variable lies between two values.
Upon substituting our given values in above formula we will get,
Upon converting 0.81859 to percentage we will get
Therefore, 81.86% of final exam score will be between 68 and 86.
Answer:
a) P ( X = 2 ) = 0.23028
b) P ( X < 4 ) = 0.95689
c) P ( X ≥ 3 | X ≥ 2 ) = 0.38292
Step-by-step explanation:
Given:-
- The parameter for the poisson distribution is given, λ = 1.3.
- Declare a random variable (X) which is the number of books sold in the next minute:
X ~ Po (1.3)
Find:-
a) What's the probability that the store will sell 2 books in the next minute? b) What's the probability that the store will sell less than 4 books in the next minute? c) What's the probability that the store will sell at least 3 books in the next minute given that it sells at least 2 books in the next minute?
Solution:-
a) The required probability P ( X = 2 ). Can be computed by using the pmf for the poisson distribution:
Where, "k" is the number of books sold in next minute.
- Now compute P ( X = 2 ) :
b) The required probability P ( X < 4 ). Can be computed by using the pmf for the poisson distribution and summing individual terms from 0 - 3:
c) The required probability P ( X ≥ 3 | X ≥ 2 ). We have to consider the conditional probability as follows:
P ( X ≥ 3 | X ≥ 2 ) = P ( X ≥ 3 & X ≥ 2 ) / P (X ≥ 2 )
= P ( X ≥ 3 ) / P (X ≥ 2 )
= P ( X > 2 ) / P ( X > 1 )
= [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]
- Compute P ( X ≤ 2 ) & P ( X ≤ 1 ) using pmf:
P ( X ≤ 2 ) = 0.27253 + 0.35429 + 0.23028
= 0.8571
P ( X ≤ 1 ) = 0.27253 + 0.35429
= 0.62682
- Use the expression developed for conditional probability to evaluate the required probability:
P ( X ≥ 3 | X ≥ 2 ) = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]
= [ 1 - 0.8571 ] / [ 1 - 0.62682 ]
= 0.38292