What is the value of x? 160° 51°
1 answer:
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Answer:
35.2% probability that the sample mean will be 246 pages or more
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using 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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
What the probability that the sample mean will be 246 pages or more?
This is 1 subtracted by the pvalue of Z when X = 246. So
By the Central Limit Theorem
has a pvalue of 0.6480.
1 - 0.6480 = 0.3520
35.2% probability that the sample mean will be 246 pages or more
Answer:
Ix = Iy =
Radius of gyration x = y =
Step-by-step explanation:
Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.
Mass of disk = ρπR2
Moment of inertia about its perpendicular axis is . Moment of inertia of quarter disk about its perpendicular is
.
Now using perpendicular axis theorem, Ix = Iy = =
.
For Radius of gyration K, equate MK2 = MR2/16, K= R/4.