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mestny [16]
3 years ago
12

1. 3х – 4 y = 24 : (0, ) ( ,0)​

Mathematics
1 answer:
kow [346]3 years ago
8 0
The points are (0,-6) (8,0)
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The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml
irina [24]

Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{120.1 - 119}{1.63}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

Z = \frac{X - \mu}{s}

Z = \frac{117.9 - 119}{1.63}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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