The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
1 answer:
Answer:
0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Also, important to remember that the standard deviation is the square root of the variance.
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 random variable X, with mean and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the probability that the sample mean would be greater than 70.5 kilograms?
This is 1 subtracted by the pvalue of Z when X = 70.5. So
By the Central limit theorem
has a pvalue of 0.7764
1 - 0.7764 = 0.2236
0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.
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Answer:
a. $1.03
b. $0.93
c. 0.98
d. 2 workers
Step-by-step explanation:
a. Given that:
- 1 computer : 1 worker : inventory 150 items per hour
- 1 computer : 2 workers : inventory 200 items per hour
- 1 computer : 3 workers : inventory 220 items per hour
- 1 computer : 4+ workers : fewer than 235 items per hour
- Cost: $100 per computer ; $25 per worker
The fixed production factor in the warehouse is the computer used:
-One computer used, but the number of users is varied to inventory a specified number of items.
-The variable production factor is the number of workers assigned per one computer.
#The cost of inventorying a single item by one worker is:
Hence, the cost of inventorying a single item is $1.03
b. Using the information provided above, the cost of inventorying a single item when two workers are assigned is :
Hence, the cost of inventorying a single item is $0.93
c.Using the information provided above, the cost of inventorying a single item when three workers are assigned is :
Hence, the cost of inventorying a single item is $0.98
d. To determine the most cost-effective job assignment, we calculate the cost of 4+ workers.
Take any number less than 235(say 234) as the inventory units:
From our calculations, it's clear that two workers per computer costs the least amount($0.93) per unit item. Hence, it is best to assign two workers per computer.