Question

**please put the solution in easy to understand letters because I also want to learn how to solve it, thanks

1) On a particular production line, the likelihood that a light bulb is defective is 10%. Seven light bulbs are randomly selected. What is the probability that at most 4 of the light bulbs will be defective?

2) Patients scheduled to see their primary care physician at a particular hospital wait, on average, an additional seven minutes after their appointment is scheduled to start. Assume the time that patients wait is exponentially distributed. What is the probability a randomly selected patient will see the doctor within eleven minutes of the scheduled time?

3) We draw a random sample of size 49 from a normal population with variance 2.1 If the sample mean is 19.5, what is a 90% confidence interval for the population mean?

a. 18.8415, 20.1585 b. 18.7180, 20.2820 c. 19,2373, 19.7627      d. 19.1600, 19.8400

4) Suppose taxi fare from Logan Airport to downtown Boston is known to be normally distributed with a standard deviation of $2.70. The last seven times John has taken a taxi from Logan to downtown Boston, the fares have been $22.50, $23.51, $23.50, $21.90, $22.42, $21.71, and $22.54. What is a 95% confidence interval for the population mean taxi fare?

5) A ramdon sample of size 36 is taken from a population with mean µ=10 and standard deviation σ =2. What is the probability that the sample mean is greater than 11?

Answer 1

Answer1) On a particular production line, the likelihood that a light bulb is defective is 10%. Seven light bulbs are randomly selected. What is the probability that at most 4 of the light bulbs will be defective?

Solution:

Given that ,

p = 0.1

1 - p = 0.9

n = 7

Using binomial distribution formula ,

P(X) = nCx * p^x (1-p)^n-x

P(Atmost 4 defective)

= P(0) + P(1) + P(2) + P(3) + P(4)

= 7C0 * 0.1^0 * 0.9^7 + 7C1 * 0.1^1 * 0.9^6 + 7C2 * 0.1^2 * 0.9^5 +7C3 *0.1^3 * 0.9^4 + 7C4 * 0.1^4 * 0.9^3

= 0.4783 + 0.3720 + 0.1240 + 0.0230 + 0.0026

= 0.9999

Therefore, the probability that at most 4 of the light bulbs will be defective is 0.9999

Answer 2) Patients scheduled to see their primary care physician at a particular hospital wait, on average, an additional seven minutes after their appointment is scheduled to start. Assume the time that patients wait is exponentially distributed. What is the probability a randomly selected patient will see the doctor within eleven minutes of the scheduled time?

Solution:

= 1/7

P(X11) = 1 - e^-(1/7*11)

= 1 - e^-1.5714

= 1 - 0.2077

= 0.7923

Answer 3) We draw a random sample of size 49 from a normal population with variance 2.1 If the sample mean is 19.5.

what is a 90% confidence interval for the population mean?

Solution:

Given that,

Sample mean, = 19.5

Standard deviation,

= √variance

= √2.1

= 1.449

n = 49

90% confidence interval for the population mean:

z-score for 90% confidence level,Z/2 = 1.645

= x̄ ± z* σ / (√n)

= 19.5 1.645​​​ * 1.449/√49

= 19.5 0.3405

= (19.1600, 19.8400)

Therefore, option d is correct answer.

Answer 4) Suppose taxi fare from Logan Airport to downtown Boston is known to be normally distributed with a standard deviation of $2.70. The last seven times John has taken a taxi from Logan to downtown Boston, the fares have been $22.50, $23.51, $23.50, $21.90, $22.42, $21.71, and $22.54. What is a 95% confidence interval for the population mean taxi fare?

Solution:

Mean, x̄ = (22.50+23.51+23.50+21.90+22.42+21.71+22.54)/7

= 158.08/7

= 22.58

Standard deviation = 2.70

95% confidence interval for the population mean taxi fare:

z-score for 95% confidence level = 1.96

= x̄ ± z* σ / (√n)

= 22.58 1.96 * 2.70/√7

= 22.58 2.00

= (20.58, 24.58)

Therefore, 95% confidence interval for the population mean taxi fare is (20.58, 24.58).

Answer 5) A ramdon sample of size 36 is taken from a population with mean µ=10 and standard deviation σ =2. What is the probability that the sample mean is greater than 11?

Solution:

n=36

µ=10

σ =2.

the probability that the sample mean is greater than 11:

P(X>11) = P((x – μ) / σ/√n > (11-10)/2/√36)

= P(Z>3)

= 1- P(Z<3)

= 1 - 0.99865

= 0.001350

~ 0.0014

Therefore, the probability that the sample mean is greater than 11 is 0.0014.