Answer:
Part 1) The volume of the ball is 256 cm³
Part 2) The radius of the ball is 3.94 cm
Step-by-step explanation:
Part 1) we know that
The density is equal to divide the mass by the volume
D=m/V
Solve for the volume
The volume is equal to divide the mass by the density
V=m/D
In this problem we have
m=128 g
D=0.5 g/cm³
substitute
V=128/0.5=256 cm³
Part 2) what is the radius of the ball?
we know that
The volume of the sphere (ball) is equal to
we have
assume
substitute and solve for r
Answer:
a) A sample of 256 was used in this survey.
b) 45.14% probability that the point estimate was within ±15 of the population mean
Step-by-step explanation:
This question is solved using the normal probability distribution and the central limit theorem.
Normal probability distribution
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.
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.
a. How large was the sample used in this survey?
We have that . We want to find n, so:
A sample of 256 was used in this survey.
b. What is the probability that the point estimate was within ±15 of the population mean?
15 is the bounds with want, 25 is the standard error. So
Z = 15/25 = 0.6 has a pvalue of 0.7257
Z = -15/25 = -0.6 has a pvalue of 0.2743
0.7257 - 0.2743 = 0.4514
45.14% probability that the point estimate was within ±15 of the population mean