Question

The mean of a population is 74 and the standard deviation is 16. The shape of the population is unknown. Determine the probability of each of the following occurring from this population. Appendix A Statistical Tables a. A random sample of size 34 yielding a sample mean of 76 or more b. A random sample of size 120 yielding a sample mean of between 73 and 75 c. A random sample of size 218 yielding a sample mean of less than 74.8

Answer 1

(a) P(
`\bar X`
`\geq` 76) = 0.2327

(b) P(73 <
`\bar X` < 75) = 0.5035

(c) P(
`\bar X` < 74.8) = 0.77035

Explanation:

We are given that the mean of a population is 74 and the standard deviation is 16.

Assuming the data follows normal distribution.

Let
`\bar X` = sample mean

The z-score probability distribution for sample mean is given by;

Z =
`(\bar X -\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = population mean = 74

`\sigma` = standard deviation = 16

n = sample size

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

(a) Probability that a random sample of size 34 yielding a sample mean of 76 or more is given by = P(
`\bar X`
`\geq` 76)

P(
`\bar X`
`\geq` 76) = P(
`(\bar X -\mu)/((\sigma)/(√(n) ) )`
`\geq`
`(76-74)/((16)/(√(34) ) )` ) = P(Z
`\geq` 0.73) = 1 - P(Z < 0.73)

= 1 - 0.7673 = 0.2327

The above probability is calculated by looking at the value of x = 0.73 in the z table which has an area of 0.7673.

(b) Probability that a random sample of size 120 yielding a sample mean of between 73 and 75 is given by = P(73 <
`\bar X` < 75) = P(
`\bar X` < 75) - P(
`\bar X`
`\leq` 73)

P(
`\bar X` < 75) = P(
`(\bar X -\mu)/((\sigma)/(√(n) ) )` <
`(75-74)/((16)/(√(120) ) )` ) = P(Z < 0.68) = 0.75175

P(
`\bar X`
`\leq` 73) = P(
`(\bar X -\mu)/((\sigma)/(√(n) ) )`
`\leq`
`(73-74)/((16)/(√(120) ) )` ) = P(Z
`\leq` -0.68) =1 - P(Z < 0.68)

= 1 - 0.75175 = 0.24825

Therefore, P(73 <
`\bar X` < 75) = 0.75175 - 0.24825 = 0.5035

The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.75175.

(c) Probability that a random sample of size 218 yielding a sample mean of less than 74.8 is given by = P(
`\bar X` < 74.8)

P(
`\bar X` < 74.8) = P(
`(\bar X -\mu)/((\sigma)/(√(n) ) )` <
`(74.8-74)/((16)/(√(218) ) )` ) = P(Z < 0.74) = 0.77035

The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.