The probability that a student gets a multiple-choice question correct on a statistics test is 0.80. Use the normal approximation to the binomial method to find the probability that a student gets a B or a C (21-26 correct answers) on a test with 30 questions. To calculate the probability, what is the upper and lower bounds of the normal approximation

Respuesta :

Answer:

0.7341 is the probability that a student gets a B or a C.

Step-by-step explanation:

We are given the following in the question:

Probability that a student gets a multiple-choice question correct  =

p = 0.80

Sample size, n = 30

We use the normal approximation to binomial distribution to solve this.

[tex]\mu = np = 30(0.8) = 24\\\sigma = \sqrt{np(1-p)} = \sqrt{30(0.8)(0.2)} = 2.19[/tex]

We have to evaluate:

a) Upper bound of marks

x = 26

[tex]P( x \leq 26) = P( z \leq \displaystyle\frac{26 - 24}{2.19}) = P(z \leq 0.9132)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x \leq 26) =0.8194[/tex]

b) Lower bound of marks

x = 21

[tex]P( x \leq 21) = P( z \leq \displaystyle\frac{21 - 24}{2.19}) = P(z \leq -1.3698)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x \leq 21) =0.0853[/tex]

c) probability that a student gets a B or a C

[tex]P(21\leq x \leq 26) = P(x < 26) - P(x < 21) = 0.8194 - 0.0853 = 0.7341[/tex]