Answer: 0.8461
Step-by-step explanation:
Given : 
Let x be the random variable that represents the cost for the hospital emergency room visit.
We assume that cost for the hospital emergency room visit is normally distributed .
z-score for x=1000 ,
![z=\dfrac{1000-1400}{392}\approx-1.02\ \ \ [\because z=\dfrac{x-\mu}{\sigma}]](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B1000-1400%7D%7B392%7D%5Capprox-1.02%5C%20%5C%20%5C%20%5B%5Cbecause%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D)
Using z-value table , we have
P-value =P(x>1000)=P(z>-1.02)=1-P(z≤ -1.02)=1-0.1538642
=0.8461358≈0.8461 [Rounded nearest 4 decimal places]
Hence, the probability that the cost will be more than $1000 = 0.8461
Answer: 186%
To get this simply divide 93 into 50 (you will get 1.86 which you can convert to the percentage 186%). An even easier way is to multiply 93 by two (186) since 50% times 2 is 100%. Hope this helps!
Step-by-step explanation:
The given sample space of S is = {ABC, ABD, ACD, BCD},
And from this sample space it can be deduced that;
(a) There can be four ways to choose the committee.
(b) There can be three ways in which person D is always included.
(c) And if persons B and C are necessarily be included, then there are two ways to form the committee.
