The total weight of Mason is 177
<em><u>Solution:</u></em>
Given that, Mr.Mason takes two-thirds of his body weight and drinks that amount of gatorade in ounces per day
Mr.Mason drinks 118 ounces each day
Let "x" be the total weight of Mr.Mason
Therefore, we can say,
Two third of body weight = Amount of gatorade drank in ounces per day
Two third of x = 118
Thus total weight of Mason is 177
Answer:
Probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.
Step-by-step explanation:
We are given that one year, professional sports players salaries averaged $1.5 million with a standard deviation of $0.9 million.
Suppose a sample of 400 major league players was taken.
<em>Let </em><em> = sample average salary</em>
The z-score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = mean salary = $1.5 million
= standard deviation = $0.9 million
n = sample of players = 400
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
So, probability that the average salary of the 400 players exceeded $1.1 million is given by = P( > $1.1 million)
P( > $1.1 million) = P(
>
) = P(Z > -8.89) = P(Z < 8.89)
<em>Now, in the z table the maximum value of which probability area is given is for critical value of x = 4.40 as 0.99999. So, we can assume that the above probability also has an area of 0.99999 or nearly close to 1.</em>
Therefore, probability that the average salary of the 400 players exceeded $1.1 million is 0.99999.