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 = ![\sqrt{npq} = \sqrt{np(1-p)} = \sqrt{58*0.3*(1-0.3)} = 3.48999](https://tex.z-dn.net/?f=%5Csqrt%7Bnpq%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B58%2A0.3%2A%281-0.3%29%7D%20%3D%203.48999)
μ = Sample mean = n·p = 58×0.3 =17.4
Therefore the probability density function is ![\frac{1}{3.49\sqrt{2 \pi } } e^{\frac{-(12-17.4)^2}{2*3.49^2} } = 3.453*10^{-2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3.49%5Csqrt%7B2%20%5Cpi%20%7D%20%7D%20e%5E%7B%5Cfrac%7B-%2812-17.4%29%5E2%7D%7B2%2A3.49%5E2%7D%20%7D%20%3D%203.453%2A10%5E%7B-2%7D)
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