Question

Question 22 A distribution of values is normal with a mean of 28.8 and a standard deviation of 20.8. Find the probability that a randomly selected value is greater than -31.5. P(X > -31.5) - Enter your answer as a number accurate to 4 decimal places. Submit Question Question 23 A distribution of values is normal with a mean of 71.9 and a standard deviation of 68. Find the probability that a randomly selected value is between - 138.9 and 24.3. P(-138.9 < X < 24.3) - Enter your answer as a number accurate to 4 decimal places. Submit Question Question 24 The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.888 g and a standard deviation of 0.302 g. Find the probability of randomly selecting a cigarette with 0.375 g of nicotine or less PIX < 0.375 g) Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted. Question Help: Video

Answer 1

Ans 23) P(X > -31.5) = 0.9981

Explanation:

Mean p = 28.8

standard deviation u = 20.8

P(X > -31.5) = 1 - P(Z < -31.5 – 28.8 20,8

ans.

/* we can find probability using excel function: =NORM.S.DIST(-2.899,TRUE) */

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Ans 24) P(-138.9 < X < 24.3) = 0.2410

Explanation:

$Mean\: \mu = 71.9$

$standard\:deviation \: \mu = 68$

$P(-138.9<X >24.3) = P(Z < \frac{24.3-71.9}{68}) - P(Z < \frac{-138.9-71.9}{68})$

ans.

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Ans 25)

Explanation:

$Mean\: \mu = 0.888$

$standard\:deviation \: \mu = 0.302$

$P(X < 0.375) = P(Z < \frac{0.375-0.888}{0.302})$

$=0.0447$ ans.