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Crazy boy [7]
3 years ago
13

What is the z score of -0.08

Mathematics
1 answer:
Ugo [173]3 years ago
3 0

Answer:

.7881

Step-by-step explanation:

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What is the inverse function of f(x)=2x-1
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Step-by-step explanation:

Given

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4 years ago
A ticket for the spring musical is $4.50. Which table show the relationship between the number of tickets purchased and the tota
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7 0
3 years ago
A common test of strength engaged in by high school students at Elmhurst HS is seeing how much weight they can lift from the flo
Alexus [3.1K]

Answer:

a) 30.85% probability that 1 randomly selected male student has the best lift less than 200 lbs.

b) 2.28% probability the sample mean will be over 245 lbs.

c) Because the underlying population(weight the students can lift) is normally distributed.

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}

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 \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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:

\mu = 225, \sigma = 50

(a) Find the probability that 1 randomly selected male student has the best lift less than 200 lbs.

This is the pvalue of Z when X = 200. So

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

Z = \frac{200 - 225}{50}

Z = -0.5

Z = -0.5 has a pvalue of 0.3085

30.85% probability that 1 randomly selected male student has the best lift less than 200 lbs.

(b) If a sample of 25 students is tested, find the probability the sample mean will be over 245 lbs.

Now n = 25, s = \frac{50}{\sqrt{25}} = 10

This probability is 1 subtracted by the pvalue of Z when X = 245. So

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

By the Central Limit Theorem

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

Z = \frac{245 - 225}{10}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability the sample mean will be over 245 lbs.

(c) Why can the normal distribution be used in part b even though the sample size is < 30?

The sample size being at least 30 condition is only if the underlying population is not normally distributed. In this case, it is, so we use the normal distribution in part b.

8 0
3 years ago
The ratio of children to adults at the football game was 23. If there were 140 children at the football game,
inessss [21]

Answer:

Therefore, the number of adults at the football game was 6.

Step-by-step explanation:

Let A represents the number of adults and C represents the number of children.

Therefore, the ratio of children to adults is given as follows:

23:1 = C:A ....................... (1)

Since C = 140, this is substituted into equation (1) as follows:

23:1 = = 140:A ....................... (2)

Transforming equation (2) and solve for A, we have:

23 / (23 + 1) = 140 / (140 + A)

23 / 24 = 140 / (140 + A)

0.958333333333333 = 140 / (140 + A)

0.958333333333333 (140 + A) = 140

(0.958333333333333 * 140) + 0.958333333333333A = 140

134.166666666667 + 0.958333333333333A = 140

0.958333333333333A = 140 -134.166666666667

0.958333333333333A = 5.833333333333

A = 5.833333333333 / 0.958333333333333

A = 6.08695652173878

Rounding to a whole number, we have:

A = 6

Therefore, the number of adults at the football game was 6.

5 0
3 years ago
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