Lecture 15/16

Probability Distributions for Continous Random Variables

Published

November 2, 2023

Section 6.3 The normal Approximation to the Binomial Probability Distribution

  • The binomial formala produces lengthy calculations and the tables are available only for certain values of n and p.
  • You can use the normal approximation to the binomial probability distribution

The normal approximation

  • Let x be a binomial random variable with n trials and probability p of success. The probability distribution of x is approximated using a normal curve with:

  • \(\mu = np\) and \(\sigma = \sqrt{npq}\).

  • This approximation is adequate as long as n is large and p is not too close to 0 or 1.

Figure 6.17

  • n = 25, p = 0.5
  • \(\mu = ?\)
  • \(\sigma = ?\)
  • compare the areas under the curve to the areas in the bar
  • Example the curve between 7.5 and 10.5 is approximately \(P(8 \le x \le 10)\)

Use Binomial Probability? \[ P(8 \le x \le 10) = .190 \]

Use Normal approximation (subtract .5 from lower end and add .5 to upper end)

\[ P(7.5 \le x \le 10.5) = .1891 \]

  • z-score

The 1/2 half adjustment is called a continuity correction and helps account for the effect that you are approximating a discrete random variable with a continuous one.

Rule of thumb:

  • The normal approximation to the binomial probabilities will be adequate if both:
  • \(np >5\) and \(nq >5\)

Figure 6.18

  • n = 25, p = .1
  • \(\mu = ?\)
  • \(\sigma = ?\)

Example 6.12

The reliability of an electrical fuse it he probability that a fuse, chosen at random from production will function under its designed conditions. A random sample of 1000 fuses was tested and x = 27 defectives were observed. Calculate the approximate probability of observing 27 or more defectives, assuming that the fuse reliability is .98.

  • \(\mu = np = 1000*.02 = 20\)

  • \(\sigma = \sqrt{npq} = \sqrt{1000 * .02 * .98} = \sqrt{19.6} = 4.43\)

  • Check if normal approximation is appropriate?

  • Create z score for value of 26.5

  • Normal approximation yields .0708

Using exact:

[1] 0.0507

Example 6.12

A soda manufacturer was fairly certain that her brand had a 10% share of the market. In a survey involving 2500 soda drinkers, x = 211 preferred her brand. If the 10% figure is correct, find the probability of observing 211 or fewer consumers who perfer her brand of soda.

  • \(\mu = 250\)

  • \(\sigma = 15\)

  • Z score for 211.5 = -2.57

\[ P(x \le 211) \approx P( z \lt -2.57) = .0051 \]

Exact binomial probability:

[1] 0.0044

What happened??














Either observed an unusual sample even though p = 0.1 OR the sample reflects the value of p is less than .1 and perhaps closer to 21/2500 = .08

Homework

[1] "6.3.13, 6.3.15, 6.3.24, 6.3.28"

Answers: Section 6.3