2.60
so <em>5</em><em> </em><em>p</em><em>o</em><em>u</em><em>n</em><em>d</em><em>s</em><em> </em><em>o</em><em>f</em><em> </em><em> </em><em>p</em><em>e</em><em>p</em><em>p</em><em>e</em><em>r</em><em> </em><em>wou</em><em>ld</em><em> </em><em>c</em><em>o</em><em>s</em><em>t</em><em> </em><em> </em><em> </em><em> </em><em>3</em><em>.</em><em>2</em><em>5</em>
Answer:
1/4
Step-by-step explanation:
The tangent line to f(x) at x=2 goes through the point (2, 8) and has slope 4.
The inverse function f⁻¹(x) is a reflection of the function f(x) across the line y=x. The corresponding tangent line will go through point (8, 2) and have slope 1/f'(2) = 1/4 at that point. That is, ...
(f⁻¹)'(8) = 1/f'(f⁻¹(8)) = 1/f'(2) = 1/4
Answer:
a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces
b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces
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 normal distributions can be 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.
Mean of 8.0 ounces and a standard deviation of 1.1 ounces.
This means that
(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?
means that
This probability is the pvalue of Z when X = 9.3. So
By the Central Limit Theorem
has a pvalue of 0.9959
0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces
(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?
means that
This probability is 1 subtracted by the pvalue of Z when X = 9. So
has a pvalue of 0.9871
1 - 0.9871 = 0.0129
0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces