Solve for the value of b

In a sample of 1000 cases, the mean of a certain test is 14 and the standard deviation is 2.5. Assume the distribution to be nor

Maslowich

Answer:

The top 20% of the students will score at least 2.1 points above the mean.

Step-by-step explanation:

When the distribution is normal, we use 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.

The mean of a certain test is 14 and the standard deviation is 2.5.

This means that $\mu = 14, \sigma = 2.5$

The top 20% of the students will score how many points above the mean

Their score is the 100 - 20 = 80th percentile, which is X when Z has a pvalue of 0.8. So X when Z = 0.84.

Their score is:

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

$0.84 = \frac{X - 14}{2.5}$

$X - 14 = 0.84*2.5$

$X = 16.1$

16.1 - 14 = 2.1

The top 20% of the students will score at least 2.1 points above the mean.

7 0

2 years ago

What is the correlation coefficient for the data shown

ivann1987 [24]

Answer:

1 is the answer

I hope this helps you

7 0

2 years ago

Write 14 is 7 times a number c as an equation

grin007 [14]

C=7•2=14 this should be right idk

7 0

3 years ago

9 buses of students, teachers and parents went on a field trip. if 5 of the buses held 63 people each and other buses held 54 pe

algol13

It is 531 maybe?????

5 0

3 years ago

What is the derivative of 1/square root 4x.

Bumek [7]

Answer:

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Exponential Properties

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg]$

Step 2: Differentiate

Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \bigg( \frac{1}{2\sqrt{x}} \bigg)'$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{\sqrt{x}} \bigg)'$
Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{x^\Big{\frac{1}{2}}} \bigg)'$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( x^\bigg{\frac{-1}{2}} \bigg)'$
Derivative Rule [Basic Power Rule]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{-1}{2} x^\bigg{\frac{-3}{2}} \bigg)$
Simplify: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4} x^\bigg{\frac{-3}{2}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0

2 years ago