Nia recorded her science and math scores. The measures of center and variation for each score are shown in the table below.

viva [34]

Marlee will have 21.05 inches of wire left, here is why;
115-25.75=89.25
89.25+30=119.25
119.25-38=81.25
81.25-60.2=21.05
She has 21.05 inches of wire left.

4 0

3 years ago

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

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.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when $X = 59.5$ . So

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

$Z = \frac{59.5 - 56}{1.12}$

$Z = 3.1$

$Z = 3.1$ has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0

3 years ago

a pice of wire 80 inches long is cut into 2 pieces. One piece is 5 inches more than twice the other. How long is the shortest pi

ad-work [718]

The longer piece of wire is 42.5

The shorter piece of wire is 37.5

37.5+42.5=80

37.5 matches the requirements because it is 5 smaller than 42.5 :)

Mark as BRAINLIEST please :)

4 0

4 years ago

Determine whether each expression is a polynomial. If it is a polynomial, state the degree of the polynomial. 3x3 - 4x2

Andrei [34K]

In this equation the degree is a polynomial.
Answer) D<span>egree 3 </span>

6 0

3 years ago

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in t

Natali5045456 [20]

Answer:

a) Kenji's time had a lower z-score, which means that he is better compared to other runners on his level, and better to his peers compared to Nedda.

b) Rachel, because her time had the lowest z-score.

Step-by-step explanation:

Z-score:

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.

In this question:

The lower the time, the better the runner is. So whoever's time has a lower z-score is a better runner. So initially, we find the z-score of each student's time.

An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in the class, ran 1 mile in 8 minutes.

This means that for Rachel's time, we have that $X = 8, \mu = 11, \sigma = 3$ . So

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

$Z = \frac{8 - 11}{3}$

$Z = -1$

A junior high school class ran 1 mile in an average of 9 minutes, with a standard deviation of 2 minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes.

This means that for Kenji's time, we have that $X = 8.5, \mu = 9, \sigma = 2$ . So

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

$Z = \frac{8.5 - 9}{2}$

$Z = -0.25$

A high school class ran 1 mile in an average of 7 minutes with a standard deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes.

This means that for Nedda's time, we have that $X = 8, \mu = 7, \sigma = 4$ . So

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

$Z = \frac{8 - 7}{4}$

$Z = 0.25$

(a) Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he?

Kenji's time had a lower z-score, which means that he is better compared to other runners on his level, and better to his peers compared to Nedda.

(b) Who is the fastest runner with respect to his or her class? Explain why

Rachel, because her time had the lowest z-score.

8 0

3 years ago