Answer:
P(2.50 < Xbar < 2.66) = 0.046
Step-by-step explanation:
We are given that Population Mean, 
= 2.58 and Standard deviation, 
= 0.75
Also, a random sample (n) of 110 households is taken.
Let Xbar = sample mean household size
The z score probability distribution for sample mean is give by;
Z = 
~ N(0,1)
So, probability that the sample mean household size is between 2.50 and 2.66 people = P(2.50 < Xbar < 2.66)
P(2.50 < Xbar < 2.66) = P(Xbar < 2.66) - P(Xbar 
2.50)
P(Xbar < 2.66) = P( < 
) = P(Z < -1.68) = 1 - P(Z 1.68)
= 1 - 0.95352 = 0.04648
P(Xbar 2.50) = P( 
) = P(Z -3.92) = 1 - P(Z < 3.92)
= 1 - 0.99996 = 0.00004
Therefore, P(2.50 < Xbar < 2.66) = 0.04648 - 0.00004 = 0.046
Idk what that answer is but good luck
Answer:
I think the Question is incorrect; the left hand side of the equation should be
m+1Cn
Step-by-step explanation:
Answer:
The distribution is
b) skewed.
The sum of the probabilities is:
1
Step-by-step explanation:
In a binomial distribution, p represents the probability of success. Success in the sense that the event of interest happens. In the model presented, the probability of success p is 0.4 since we are informed that 40% of adults watch a particular television show.
The next quantity of significance in a binomial model is the number of independent trials, n. In our case there are 6 independent trials since we are told that 6 adults were selected at random. If we let the random variable K denote the number of adults out of the 6 who watch the television show, then K is a binomial random variable with parameters;
n = 6 and p = 0.4
A binomial distribution is only symmetric when either p is 0.5 or n is large. In the presented scenario none of this conditions is met since p is 0.4 while n is just 6 which is relatively small. Thus we conclude that the distribution is not symmetric but rather skewed.
The sum of the probabilities is any discrete probability distribution such as the bernoulli, binomial, negative binomial, poisson, or the geometric distribution is always equal to 1. That's a rule of thumb.
