X* square root of 3 =34

The College Boards, which are administered each year to many thousands of high school students, are scored so as to yield a mean

Marysya12 [62]

Answer:

a) 15.87% of the scores are expected to be greater than 600.

b) 2.28% of the scores are expected to be greater than 700.

c) 30.85% of the scores are expected to be less than 450.

d) 53.28% of the scores are expected to be between 450 and 600.

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 500, \sigma = 100$

a. Greater than 600

This is 1 subtracted by the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 500}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

1 - 0.8413 = 0.1587

15.87% of the scores are expected to be greater than 600.

b. Greater than 700

This is 1 subtracted by the pvalue of Z when X = 700. So

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

$Z = \frac{700 - 500}{100}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% of the scores are expected to be greater than 700.

c. Less than 450

Pvalue of Z when X = 450. So

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

$Z = \frac{450 - 500}{100}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085.

30.85% of the scores are expected to be less than 450.

d. Between 450 and 600

pvalue of Z when X = 600 subtracted by the pvalue of Z when X = 450. So

X = 600

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

$Z = \frac{600 - 500}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 450

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

$Z = \frac{450 - 500}{100}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085.

0.8413 - 0.3085 = 0.5328

53.28% of the scores are expected to be between 450 and 600.

6 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span> $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$