Let be the weight of i-th player.
1. If the mean weight of 4 backfield members on the football team is 221 lb, then
2. If the mean weight of the 7 other players is 202 lb, then
3. From the previous statements you have that
Add these two equalities and then divide by 11:
Answer: the mean weight of the 11-person team is
Answer:
Step-by-step explanation:
Given
Required
Determine side x
The hypotenuse (H) is calculated using:
Substitute 9 for x:
So:
Answer:
24/49
Step-by-step explanation:
Answer:
0.007%
Step-by-step explanation:
4000/28=0.007%
Answer:
You should expect to find the middle 98% of most head breadths between 3.34 in and 8.46 in.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean and standard deviation , the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question:
In what range would you expect to find the middle 98% of most head breadths?
From the: 50 - (98/2) = 1st percentile.
To the: 50 + (98/2) = 99th percentile.
1st percentile:
X when Z has a pvalue of 0.01. So X when Z = -2.327.
99th percentile:
X when Z has a pvalue of 0.99. So X when Z = 2.327.
You should expect to find the middle 98% of most head breadths between 3.34 in and 8.46 in.