First, you need to put them in order
53kg, 55kg, 61kg, 61kg, 76kg, 91kg, 98kg, 105kg, 120kg
For mean, you add them all up and divide by the amount of numbers (9)
720/9 = 80
For median, you find the middle number (76kg)
For mode, you find the number that appears the most (61kg)
Mean: 80
Median: 76
Mode: 61
Answer:
the probability that a sample of the 35 exams will have a mean score of 518 or more is <em> 0.934 </em>or<em> 93.4%</em>.
Step-by-step explanation:
This is s z-test because we have been given a sample that is large (greater than 30) and also a standard deviation. The z-test compares sample results and normal distributions. Therefore, the z-statistic is:
(520 - 518) / (180/√35)
= 0.0657
Therefore, the probability is:
P(X ≥ 0.0657) = 1 - P(X < 0.0657)
where
- X is the value to be standardised
Thus,
P(X ≥ 0.0657) = 1 - (520 - 518) / (180/√35)
= 1 - 0.0657
= 0.934
Therefore, the probability that a sample of the 35 exams will have a mean score of 518 or more is <em>0.934 or 93.4%</em>.
Answer: 0.5898
Step-by-step explanation:
Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .
We assume that,
The probability that .J. Redick makes any given free throw =0.901 (1)
Free throws are independent.
So it is a binomial distribution .
Using binomial probability formula, the probability of getting success in x trials :
, where n= total trials
p= probability of getting in each trial.
Let x be binomial variable that represents the number of a=makes.
n= 14
p= 0.901 (from (1))
The probability that he makes at least 13 of them will be :-
![=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898](https://tex.z-dn.net/?f=%3D%5E%7B14%7DC_%7B13%7D%280.901%29%5E%7B13%7D%281-0.901%29%5E1%2B%5E%7B14%7DC_%7B14%7D%280.901%29%5E%7B14%7D%281-0.901%29%5E0%5C%5C%5C%5C%3D%2814%29%280.901%29%5E%7B13%7D%280.099%29%2B%281%29%280.901%29%5E%7B14%7D%5C%20%5C%20%5B%5Cbecause%5C%20%5EnC_n%3D1%5C%20%5C%26%5C%20%5EnC_%7Bn-1%7D%3Dn%20%5D%5C%5C%5C%5C%5Capprox0.3574%2B0.2324%3D0.5898)
∴ The required probability = 0.5898
Answer:
7
3
all real numbers
y>0
Step-by-step explanation:
Just got it right on edge. i got you little bro
