The distance be Find the distance between P and Q
1 answer:
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Answer:
e. 0.08
Step-by-step explanation:
In the question above, a certain quantity of goods was supplied while a specific quantity of goods was sold per week. In a given week, if the number of proportion sold is X, therefore:
f(x) = {Γ(4+2)/Γ(4)Γ(2) x^3 (1-x), 0≤x≤1 ; 0, elsewhere
and
P(X greater than 0.9) = = 20*{(y^4/4)[1,0.9] - (y^5/5)[1,0.9]} = 20*{(0.25 - 0.164) - (0.20 - 0.118)} = 20*{0.086 - 0.0819} = 20*0.0041 = 0.082
Therefore the probability of the proportion sold is approximately 0.082
Answer:
The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For the population, we have that:
Mean = 15
Standard deviaiton = 12
Sample of 30
By the Central Limit Theorem
Mean 15
Standard deviation
Approximately normal
The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.