Question

Use the normal distribution to approximate the desired probability. A coin is tossed 22 times. A person, who claims to have extrasensory perception, is asked to predict the outcome of each flip in advance. She predicts correctly on 13 tosses. What is the probability of being correct 13 or more times by guessing?
A)
26.57156%
B)
25.67156%
C)
26.42156%
D)
26.37156%
E)
25.95490%
F)
26.12156%
G) None of These

Answer 1

Step-by-step explanation:

Let X be random variable defined as the number of tosses predicted correctly. Then X follows binomial distribution with parameters n and p  

The information provided is:

n = number of tosses = 22

X = number tosses predicted correctly = 13

p = probability of success = 0.5

$\hat{p}$ = sample probability of success = $\frac{X}{n}=\frac{13}{22}=0.591$

Normal Approximation to binomial is used to solve this problem since the conditions $np>10\ and\ n(1-p) >10$ are satisfied.

• $np=22\times0.5 =11>10$

• $n(1-p)=22\times(1-0.5)=11>10$

Compute the value of $P(X\geq 13)$ as follows:

$P(X\geq 13)=P(Z\geq z)$

Compute the z-score:

$z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p}{n}}}\\=\frac{0.591-0.5}{\sqrt{\frac{0.5(1-0.5)}{22}}}\\=0.854$

Use the Normal Distribution table to compute the probability.

$P(X\geq 13)=P(Z\geq 0.854)\\=1-P(Z<z)\\=1-0.8023\\=0.1977$

Thus, the probability of being correct 13 or more times by guessing is 19.77%.

The correct option is (G).