Finding Bone Density Scores. In Exercises 37–40 assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the bone density test score corresponding to the given information. Round results to two decimal places. 38. Find P 10 , the 10th percentile. This is the bone density score separating the bottom 10% from the top 90%.
Finding Bone Density Scores. In Exercises 37–40 assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the bone density test score corresponding to the given information. Round results to two decimal places. 38. Find P 10 , the 10th percentile. This is the bone density score separating the bottom 10% from the top 90%.
Finding Bone Density Scores. In Exercises 37–40 assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the bone density test score corresponding to the given information. Round results to two decimal places.
38. Find P10, the 10th percentile. This is the bone density score separating the bottom 10% from the top 90%.
Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.
The masses measured on a population of 100 animals were grouped in the
following table, after being recorded to the nearest gram
Mass
89 90-109 110-129 130-149 150-169 170-189 > 190
Frequency 3
7 34
43
10
2
1
You are given that the sample mean of the data is 131.5 and the sample
standard deviation is 20.0. Test the hypothesis that the distribution of masses
follows a normal distribution at the 5% significance level.
state without proof
the uniqueness
theorm of probability
function
(a+b)
R2L
2+2*0=?
Ma
state without proof the uniqueness theorm
of probability function suppose thatPandQ
are probability measures defined on the
same probability space (Q, F)and that
Fis generated by a π-system if P(A)=Q(A)
tax for all A EthenP=Q i. e. P(A)=Q(A) for alla g
// معدلة 2:23 ص
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