t value at 95% = 2.0639
CI = 172.5+ /- 2.0639 *(15.4/sqrt(25))
= (166.1432,178.8568)
b)
yes
c)
P(x < 178.86)
= P(z < (178.86 - 172.5)/(15.4/sqrt(25))
= P(z < 2.0649)
= 0.9805
d)
P(x < 166.14)
= P(z < (166.14 - 172.5)/(15.4/sqrt(25))
= P(z < -2.0649)
= 0.0195
e)
P(x = 172.5)
= P(z = (172.5- 172.5)/(15.4/sqrt(25))
= P(z =0)
= 0.05
a random sample of 25 is collected for a continuoue random variable x. the observed sample...
A simple random sample of size nequals=25 is drawn from a population that is normally distributed. The sample mean is found to be x overbar equals 65 and the sample standard deviation is found to be s equals=20. Construct a 90% confidence interval about the population mean.
A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found to be 108, and the sample standard deviation, s, is found to be 10. (a) Construct a 95% confidence interval about mu if the sample size, n, is 25. (b) Construct a 95% confidence interval about mu if the sample size, n, is 12. (c) Construct a 70% confidence interval about mu if the sample size, n,...
1. Let X be a normal random variable with mean 16. If P(X < 20) 0.65, find the standard deviation o. 2. The probability that an electronic component will fail in performance is 0.2 Use the normal approximation to Binomial to find the probability that among 100 such components, (a) at least 23 will fail in performance. X 26) (b) between 18 and 26 (inclusive) will fail in performance. That is find P(18 3. If two random variables X and...
2. A random sample of 25 watermelons from New Seasons were weighed, generating a sample mean of xã = 21 pounds with a standard deviation of s = 3 pounds. a) Construct a 98% confidence interval for the population mean weight of watermelons at New Seasons, b) Interpret: c) New Seasons claims that their watermelons have an average weight of 22 pounds. Does your sample from above provide sufficient evidence that the population mean of watermelons is actually less than...
10. A simple random sample of size n is drawn. The sample mean x is found to be 39.1, and the sample standard deviation s is found to be 9.7. a) (2 points) Construct a 90% confidence interval for the population mean u if the sample size n is 41. b) (2 points) Construct a 90% confidence interval for the population mean y if the sample size n is 101. c) (2 points) Construct a 99% confidence interval for the...
10. A simple random sample of size n is drawn. The sample mean x is found to be 39.1, and the sample standard deviation s is found to be 9.7. a) (2 points) Construct a 90% confidence interval for the population mean w if the sample size n is 41. b) (2 points) Construct a 90% confidence interval for the population mean 4 if the sample size n is 101. c) (2 points) Construct a 99% confidence interval for the...
The population s1700. A random sample of 25 managers is selected from this population. Find the probability that the mean annual salary of the sample is less than $55,000 mean annual salary for managers is $59,100, with a standard deviation of The population s1700. A random sample of 25 managers is selected from this population. Find the probability that the mean annual salary of the sample is less than $55,000 mean annual salary for managers is $59,100, with a standard...
In a random sample of 25 people, the mean commute time to work was 32.9 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a t-distribution to construct a 98% confidence interval for the population mean μ: What is the margin of error of μ? Interpret the results. The confidence interval for the population mean μ is _______ The margin of error of μ is _______ Interpret the results A. If a large sample of people are...
Suppose a random sample of n = 25 observations is selected from a population that is normally distributed with mean equal to 109 and standard deviation equal to 15. (a) Find the probability that x exceeds 113. (b) Find the probability that the sample mean deviates from the population mean μ = 109 by no more than 5.
4. A simple random sample of 100 customers is selected from an account receivable portfolio and the sample mean account balance is $3000. The population standard deviation σ is known to be $300. ( 8 marks) a. Construct a 90% confidence interval for the mean account balance of the population. b. What is the margin of error for estimating the mean account balance at the 90% confidence level? c. Construct a 90% confidence interval for the mean account balance of...