Question

Compute​ P(x) using the binomial probability formula. Then determine whether the normal distribution can be used to estimate this probability. If​ so, approximate​ P(x) using the normal distribution and compare the result with the exact probability.A) For n=58, p= 0.3, and x=12, use the binomial probability formula to find P(X).B) Approxiamte P(x) using the normal distribution. Use a standard distribution table.C) By how much to the probabilities differ?D) Why can the normal distrubution be used to approximate this probabilitya. because np(1-p) %u2265 10b. because np(1-p) %u2264 10c. because %u221Anp(1-p) %u2265 10d. because %u221Anp(1-p) %u2264 10E) When contracting a normal probability plot, what does fi represent?a. the value of fi represents the expected number of observation less than or equal to the ith data valueb. the value of fi represents the expectedz-score of the ith data valuec. the value of fi represents the expected proportion of observation less than or equal to the ith data value

Answer 1

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 $z = \frac{x-\mu}{\sigma}$ = -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

$\frac{1}{\sigma\sqrt{2 \pi } } e^{\frac{-(x-\mu)^2}{2\sigma^2} }$

where:

σ = Standard deviation = $\sqrt{npq} = \sqrt{np(1-p)} = \sqrt{58*0.3*(1-0.3)} = 3.48999$

μ = Sample mean = n·p = 58×0.3 =17.4

Therefore the probability density function is $\frac{1}{3.49\sqrt{2 \pi } } e^{\frac{-(12-17.4)^2}{2*3.49^2} } = 3.453*10^{-2}$

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$_i$ 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