Answer:
There are 118 plants that weight between 13 and 16 pounds
Step-by-step explanation:
For any normal random variable X with mean μ and standard deviation σ : X ~ Normal(μ, σ)
This can be translated into standard normal units by :
Let X be the weight of the plant
X ~ Normal( 15 , 1.75 )
To find : P( 13 < X < 16 )
= P( -1.142857 < Z < 0.5714286 )
= P( Z < 0.5714286 ) - P( Z < -1.142857 )
= 0.7161454 - 0.1265490
= 0.5895965
So, the probability that any one of the plants weights between 13 and 16 pounds is 0.5895965
Hence, The expected number of plants out of 200 that will weight between 13 and 16 = 0.5895965 × 200
= 117.9193
Therefore, There are 118 plants that weight between 13 and 16 pounds.
Answer:
5,5,7,*9,11,*12,16,23
Step-by-step explanation:
Since, the data has an even amount of numbers once, you put your numbers in order you place your pointer fingers from the 1st to last number & continue to move closer. There is an even amount of numbers so you would have to add the 2 numbers up & then divide them by 2.
9+11=20 divided by 2=10
Answer:
1740 is the time on a 24 hour clock or military time
Answer and Explanation:
Given : The random variable x has the following probability distribution.
To find :
a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution.
b. Calculate the expected value of x.
c. Calculate the variance of x.
d. Calculate the standard deviation of x.
Solution :
First we create the table as per requirements,
x P(x) xP(x) x² x²P(x)
0 0.25 0 0 0
1 0.20 0.20 1 0.20
2 0.15 0.3 4 0.6
3 0.30 0.9 9 2.7
4 0.10 0.4 16 1.6
∑P(x)=1 ∑xP(x)=1.8 ∑x²P(x)=5.1
a) To determine that table shows a probability distribution we add up all five probabilities if the sum is 1 then it is a valid distribution.
Yes it is a probability distribution.
b) The expected value of x is defined as
c) The variance of x is defined as
d) The standard deviation of x is defined as
