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Suppose babies born in a large hospital have a mean weight of 4090 grams, and a variance of 313,600. If 64 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by greater than 43 grams

2 Answers

5 votes

Final answer:

To find the probability that the mean weight of the sample babies would differ from the population mean by greater than 43 grams, we need to calculate the z-score and use a z-table or calculator to find the probability.

Step-by-step explanation:

To solve this problem, we first need to calculate the standard deviation of the sample mean. The standard deviation of the sample mean is equal to the standard deviation of the population divided by the square root of the sample size. In this case, the standard deviation of the population is the square root of the variance, which is 560.

The standard deviation of the sample mean is therefore 560 divided by the square root of 64, which is 70.

We can then calculate the z-score of the difference between the sample mean and the population mean. The z-score is equal to the difference between the sample mean and the population mean, divided by the standard deviation of the sample mean.

In this case, the z-score is equal to 43 divided by 70, which is approximately 0.614.

We can then use a z-table or a calculator to find the probability that a z-score is greater than 0.614. The probability is the area under the standard normal curve to the right of the z-score. Using a z-table or a calculator, we find that the probability is approximately 0.267.

User Atrash
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6.7k points
5 votes

Answer:

Step-by-step explanation:

mean weight of baby = 4090 gms

standard deviation = √ 313600

= 560

z value = deviation / standard deviation

= 43 / 560

= .0767

User Akohout
by
7.2k points