Answer to Question #214813 in Statistics and Probability for avis

Question

Suppose a survey by the Department of Roads in 2019 revealed that 60% of the vehicles travelling on E to F Highway, where speed limits are posted at 80 kilometres per hour are found to be exceeding the limit. Suppose you randomly record the speed of 10 vehicles travelling on the highway as part of your internship. The sample size of the entire survey is not disclosed to you.

Compute the following probabilities. Showing all working.

a) P (X = 2).

b) P (X = 5).

c) P (X = 10).

A random sample of high school students are asked the number of hours they spent on social media. The hours are reported below: 3 5 6 7 8 10 12 13 14 15 You can assume the times are normally distributed with a standard deviation of 8.5.

a) Estimate with 90 percent confidence the mean number of hours spent on social media by all high school students. Show all working as part of your answer and interpret your answer.

Answer 1

1. $X\sim Bin (n, p)$

$$P(X=x)=\dbinom{n}{x}p^x(1-p)^{n-x}$$

Given $n=10, p=0.6$

a)

$$P(X=2)=\dbinom{10}{2}(0.6)^2(1-0.6)^{10-2}$$

$$=0.010616832$$

b)

$$P(X=5)=\dbinom{10}{5}(0.6)^5(1-0.6)^{10-5}$$

$$=0.2006581248$$

c)

$$P(X=10)=\dbinom{10}{10}(0.6)^{10}(1-0.6)^{10-10}$$

$$=0.0060466176$$

2.

$$mean=\bar{x}=\dfrac{1}{10}(3 +5+ 6+ 7 +8$$

$$+ 10 +12 +13 +14 +15)=9.3$$

The critical value for $\alpha=0.1$ is $z_c=z_{1-\alpha/2}=1.6449.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(9.3-1.6449\times\dfrac{8.5}{10}, 9.3+1.6449\times\dfrac{8.5}{10})$$

$$=(4.8786, 13.7214)$$

Therefore, based on the data provided, the 90% confidence interval for the population mean is $4.8786<\mu<13.7214,$ which indicates that we are 90% confident that the true population mean $\mu$ is contained by the interval $(4.8786, 13.7214).$