Question

question #6
for a population, the mean is $360 with a standard deviation of $87. determine the probability of a sample of 12 having a mean of $400 or higher.
.2617
.0556
.1837
.3228

question #7
a random sample of size n = 16 is to be taken from a normal population having a mean of 100 and standard deviation of 16. determine the probability that the sample mean is greater than 108.
.6915
.3085
.0228
.9772

Explanation:

Step1: Calculate the standard error

The formula for standard error \(SE=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 87\), \(n = 12\), then \(SE=\frac{87}{\sqrt{12}}\approx25.1\)

Step2: Calculate the z - score

The formula for z - score \(z=\frac{\bar{x}-\mu}{SE}\). Given \(\bar{x} = 400\), \(\mu=360\), \(SE\approx25.1\), then \(z=\frac{400 - 360}{25.1}\approx1.59\)

Step3: Find the probability

Using the standard normal distribution table, \(P(Z\geq1.59)=1 - P(Z < 1.59)\). From the table, \(P(Z < 1.59)=0.9441\), so \(P(Z\geq1.59)=1 - 0.9441 = 0.0559\approx0.0556\)

Step1: Calculate the standard error

The formula for standard error \(SE=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 16\), \(n = 16\), then \(SE=\frac{16}{\sqrt{16}} = 4\)

Step2: Calculate the z - score

The formula for z - score \(z=\frac{\bar{x}-\mu}{SE}\). Given \(\bar{x} = 108\), \(\mu = 100\), \(SE = 4\), then \(z=\frac{108-100}{4}=2\)

Step3: Find the probability

Using the standard normal distribution table, \(P(Z\geq2)=1 - P(Z < 2)\). From the table, \(P(Z < 2)=0.9772\), so \(P(Z\geq2)=1 - 0.9772=0.0228\)

Answer:

.0556