The cost of 5 gallons of ice cream has a variance of 49 with a mean of 37 dollars during the summer. What is the probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected? Round your answer to four decimal places.

0.0119 = 1.19% probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected

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(which is the square root of the variance) , 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 problem, we have that:

What is the probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 35.2.

By the Central Limit Theorem

has a pvalue of 0.0119

0.0119 = 1.19% probability that the sample mean would be less than 35.2 dollars if a sample of 77 5-gallon pails is randomly selected

exactly you go it

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)