Question

Suppose Siri (an Apple digital assistant) is known to answer general facts correctly approximately 44.52% of the time. If a consumer asked Siri 181 questions about general facts, what is the probability that less than 45.1% of them would be correct

Answer 1

Answer:

The probability that there will be less than 45.1% questions of general facts that Siri answers correctly is 0.5636.

Step-by-step explanation:

Let X = number of times Siri is known to answer general facts correctly.

The probability of random variable X is, P (X) = p = 0.4452.

The sample selected is of size, n = 181.

The random variable $X\sim Bin(181, 0.4452)$

As the sample size is large, i.e. n > 30 and the [probability of success is closer to 0.50, i.e.p is close to 0.50, then the binomial distribution can be approximated by the normal distribution.

Also if np ≥ 10 and n (1 - p) ≥ 10 then binomial distribution can be approximated by normal distribution.

Check the conditions as follows:

• n = 181 > 30
• p = 0.4452 is closer to 0.50
• np = 181 × 0.4452 = 80.5812 > 10
• n (1 - p) = 181 × (1 - 0.4452) = 100.4188 > 10

All the conditions are satisfied.

Then the sample proportion ($\hat p$) follows a Normal distribution.

Mean = $\mu_{\hat p}=np=181\times0.4452=80.5812\approx80.6$

Standard deviation = $\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.4452\times(1-0.4452)}{181}} =0.037$

Compute the probability that there will be less than 45.1% questions of general facts that Siri answers is correct as follows:

$P(\hat p<0.451)=P(\frac{\hat p-\mu_{\hat p}}{\sigma{\hat p}}<\frac{0.451-0.4452}{0.037})=P(Z<0.16)=0.5636$

**Use the z-table for probability.

Thus, the probability that there will be less than 45.1% questions of general facts that Siri answers correctly is 0.5636.