In the Jefferson Middle School chorus, 60% of the members are girls. If there are 45 girls in chorus what is the total number of
1 answer:
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We will use the following law of indices (or 'index law') to check each pair of expression
![x^{ \frac{m}{n}} = ( \sqrt[n]{x} )^{m}](https://tex.z-dn.net/?f=%20x%5E%7B%20%5Cfrac%7Bm%7D%7Bn%7D%7D%20%3D%20%28%20%5Csqrt%5Bn%5D%7Bx%7D%20%29%5E%7Bm%7D%20)
With fractional power, the denominator is the root and the numerator is the power of the term. When the denominator is 2, we usually only write the normal square symbol (√). Denominator other than 2, we usually write the value of the root, for example, the cubic root ∛
Option A - Incorrect
should equal to ![( \sqrt[7]{7} )^{5}](https://tex.z-dn.net/?f=%20%28%20%5Csqrt%5B7%5D%7B7%7D%20%29%5E%7B5%7D%20)
Option B - Correct
does equal to 
Option C - Incorrect
should equal to 
Option D - Incorrect
should equal to ![( \sqrt[3]{5} )^{2}](https://tex.z-dn.net/?f=%20%28%20%5Csqrt%5B3%5D%7B5%7D%20%29%5E%7B2%7D%20)
With fractional power, the denominator is the root and the numerator is the power of the term. When the denominator is 2, we usually only write the normal square symbol (√). Denominator other than 2, we usually write the value of the root, for example, the cubic root ∛
Option A - Incorrect
Option B - Correct
Option C - Incorrect
Option D - Incorrect
Answer:
The answers to the questions are;
A) P(X) where x = 12 is equal to 3.55 ×10⁻²
B) P(x) using the normal distribution is given by the probability distribution function and is equal to 3.453 ×10⁻²
C) The calculated probabilities of P(x=12) using the binomial probability formula and the probability distribution function differ by 9.7×10⁻⁴
D) The normal distribution be used to approximate this probability because a. because np(1-p) %u2265 ⇒ np(1-p) ≥ 10
E) c. the value of fi represents the expected proportion of observation less than or equal to the ith data value.
Step-by-step explanation:
To solve the question, we note that
n = 58
p = 0.3
x = 12
Here we have n·p = 17.4 and n·q = 40.6 both ≥ 5 that is the normal distribution can be used to estimate the probability
A) We are to use the binomial probability formula to find P(X)
Where x = 12, we have P(12) = ₅₈C₁₂×0.3¹²×0.7⁴⁶
= 891794789340×0.000000531441×7.49×10⁻⁸ = 3.55 ×10⁻²
B) Using the standard distribution table we have
the z score for x = 12 given as = -1.55
From the normal distribution table, we have the probability that value is below 12 = .06057 while the normal probability distribution function which gives the probability of a number being 12 is given by
where:
σ = Standard deviation =
μ = Sample mean = n·p = 58×0.3 =17.4
Therefore the probability density function is
C) The probabilities differ by 3.55 ×10⁻² - 3.453 ×10⁻² = 9.7×10⁻⁴
D) The normal distribution be used to approximate this probability because n·p and n·p·(n-p) ≥ 5
E) f represents the area under the curve towards left of the ith data observed in a normally distributed population.
Therefore, the value of fi represents the expected proportion of observation less than or equal to the ith data value