Answer:
The answer is "Option a".
Step-by-step explanation:
[tex]n= 93 \\\\p= 0.24\\\\\mu=?\\\\ \sigma=?\\\\[/tex]
Using the binomial distribution: [tex]\mu = n\times p = 93 \times 0.24 = 22.32\\\\\sigma = \sqrt{n \times p \times (1-p)}=\sqrt{93 \times 0.24 \times (1-0.24)}=4.1186[/tex]
In this the maximum value which is significantly low, [tex]\mu-2\sigma[/tex], and the minimum value which is significantly high, [tex]\mu+2\sigma[/tex], that is equal to:
[tex]\mu-2\sigma = 22.32 - 2(4.1186) = 14.0828 \approx 14.08\\\\\mu+2\sigma = 22.32 + 2(4.1186) = 30.5572 \approx 30.56[/tex]
