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<span>Given that there are 50 campers at day camp and 10
counselors.
The ratio of campers to counselors as a fraction is given by
.
Using equivalent fractions we have
.
Thus, the related rate and unit rate to the ratio of campers to counselors is given by
.
The unit rate indicates that for every counselor, there are 5 campers.</span>
The ratio of campers to counselors as a fraction is given by
Using equivalent fractions we have
Thus, the related rate and unit rate to the ratio of campers to counselors is given by
The unit rate indicates that for every counselor, there are 5 campers.</span>
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.