Answer:
a. Mean = 6
Variance = 2.4
Standard Deviation = 1.55
b. P(X=16) = 0.124
Step-by-step explanation:
Given
n = Total shots = 10
p = Probability of success = 60%
p = 60/100
p= 0.6
q = Probability of failure
q = 1-p
q = 1 - 0.6
q = 0.4
a.
Mean = np
Mean = 10 * 0.6
Mean = 6
Variance = npq
Variance = 10 * 0.6 * 0.4
Variance = 2.4
Standard Deviation = √Variance
Standard Deviation = √2.4
Standard Deviation = 1.549193338482966
Standard Deviation = 1.55 --------- approximated
b.
We have X = 16
x = 10
Assume that the events "success" on the various throws are independent.
The 10th success came on the 16th attempt
So, the player had exactly 10 successes and 6 failures on 16th trial
So Probability = nCr 0.6^10 * 0.4^6
Where n = 15 and r = 9 (number of attempts and success before the 16th trial)
15C9 * 0.6^10 * 0.4^6
= 5005 * 0.0060466176 * 0.004096
= 0.123958563176448
= 0.124 ------ Approximated
You could separate the equation:
2*50+ 2*3=
I know that 2*50= 100; 2*3= 6
100+6= 106
I hope this helps!
~cupcake
Answer: I don't see the amount per football
Step-by-step explanation:
