Mathematics, 22.04.2020 01:10 katherineedwards1105

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

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Answer from: emma6892

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

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

Step-by-step 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 = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ 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( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{76-74}{\frac{16}{\sqrt{34} } }$ ) = 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( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{75-74}{\frac{16}{\sqrt{120} } }$ ) = P(Z < 0.68) = 0.75175

P( $\bar X$ $\leq$ 73) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{73-74}{\frac{16}{\sqrt{120} } }$ ) = 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( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{74.8-74}{\frac{16}{\sqrt{218} } }$ ) = 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.

Answer from: Quest

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Segment ab is dilated from the origin to create segment a prime b prime at a' (0, 6) and b' (6, 9).

Answer from: Quest

You just divide and you will find x

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The mean of a population is 74 and the standard deviation is 16. The shape of the population is unkn...