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]
