A random variable

following a binomial distribution with success probability

across

trials has PMF

where

.
The mean of the distribution is given by the expected value which is defined by

where the summation is carried out over the support of

. So the mean is

Because this is a proper distribution, you have

which is a fact that will be used to evaluate the sum above.





Letting

, this becomes

Observe that the remaining sum corresponds to the PMF of a new random variable

which also follows a binomial distribution with success probability

, but this time across

trials. Therefore the sum evaluates to 1, and you're left with

as the expression for the mean for

.
Answer:
a. 5.26
Step-by-step explanation:
As, the standard deviation of the distribution of the difference between mean is 0.76 so the t score will be simply calculated by dividing the difference of means by standard deviation of the distribution of the difference between mean as hypothesized difference is zero.
t score=Xbar A- Xbar B/Sd of difference of means
t score=55-51/0.76
t score=4/0.76=5.26
B U R G E R = 6 letters
S T A R B I R D = 8 letters
14 letters in all
B = 2 ; U = 1 ; R = 4 ; G = 1 ; E = 1 ; S = 1 ; T = 1 ; A = 1 ; I = 1 ; D = 1
Probability of pulling out an R: 4/14
Probability of pulling out a B: 2/14
Probability of pulling out a letter that appears in the first half of the alphabet: 7/14
Probability of pulling a vowel: 4/14
Answer:
84miles
Step-by-step explanation:
Step one:
given data
we are told that the equation that describes the car's speed is
d = 70t-----------1
<em>From the equation, we can see that the speed with which the car is traveling is 70miles per hour, while we are expected to find the distance d when the time is 1.2 hours.</em>
<em />
Required
The distance d at 1.2 hours
Step two:
substituting in the expression we have
d= 70*1.2
d=84miles
7.24/7=1.04
7.00/7= 1 0.24/7= .04 1+0.04=1.04
11.4/19= 0.6
https://www.calculatorsoup.com/calculators/math/longdivisiondecimals.php?dvsor=19&dvdnd=11.4&decimal...
Kinda hard to show work for second one so posted a link.