Question

A normally distributed population has a mean of 500 and a standard deviation of 80. a. Determine the probability that a random sample of size 25 selected from this population will have a sample mean less than 463 . b. Determine the probability that a random sample of size 16 selected from the population will have a sample mean greater than or equal to 538.

A company makes windows for use in homes and commercial buildings. The standards for glass thickness call for the glass to average 0.350 inch with a standard deviation equal to 0.070 inch. Suppose a random sample of n=58 windows yields a sample mean of 0.360 inch. Complete parts a and b below.

a. What is the probability of x≥0.360 if the windows meet the​ standards?

Answer 1

Mean p = 500

Standard deviation o = 80

Formula to calculate z-score:

$z = \frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}$

Ans a)

$P(X < 463) = P(Z < \frac{463-500}{\frac{80}{\sqrt{25}}})$

ans.

/* we can find probability using excel function: =NORM.S.DIST(-2.31,TRUE) */

Probability population will have a sample mean less than 463 = 0.0104

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Ans b)

$P(X \geq538) = 1 - P(Z < \frac{538-500}{\frac{80}{\sqrt{16}}})$

ans.

Probability sample mean greater than or equal to 538 = 0.0287

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$Mean \: \mu =0.350$

$Standard\: deviation \: \sigma =0.070$

Ans a)

$P(X \geq0.360) = 1 - P(Z < \frac {0.360-0.350}{\frac{0.070}{\sqrt{58}}})$

$= 1 - P(Z < \frac {0.360-0.350}{\frac{0.070}{\sqrt{58}}})$

$=0.1379$ ans.

Therefore, Probability is 0.1379