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charle [14.2K]
3 years ago
11

PLS HELP GIVING BRAINLIEST FOR THE CORRECT ANSWER (NEEDS ANSWER IN 5 MINUTES OR ELSE NO BRAINLIEST)

Mathematics
1 answer:
Umnica [9.8K]3 years ago
3 0

Answer:

9

Step-by-step explanation:

99/2 × (11+991) = 49,599

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JIM TOSSES A BALL IN THE AIR WHEN ON THE COUNT OF 2, 3 , 5 8 13 what numbers will he count when he tosses the ball for the 6th 7
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Answer:

21, 34, and 55 because it's like that one ratio, I forgot the name tho

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Among 39- to 44-year-olds, 35​% say they have called a talk show while under the influence of peer pressure. Suppose seven 39 to
LenKa [72]
<span>if 35% or 0.35 say no, to compute the probability that at least one of 7 </span>called a talk show while under the influence of peer pressure <span> is (1-0.35)^7  and subtract it from . This is the same as at least 1 has </span><span>called a talk show while under the influence of peer pressure
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3 years ago
The mean life of a television set is 119119 months with a standard deviation of 1414 months. If a sample of 7474 televisions is
Pepsi [2]

Answer:

0.5034 = 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 can be 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}

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 \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 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}{1.6275}

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}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

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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Answer:

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