Question 510932: Ok, These 6 problems I apparently did wrong on my test... Can someone do them and explain them a bit please?
1. A set of 50 data values has a mean of 28 and a variance of 4.
a. Find the standard score (z) for a data value = 31.
b. Find the probability of a data value > 31.
2.Find the area under the standard normal curve:
a. to the right of z = 2.37
b. to the left of z = 2.37
3. Assume that the population of heights of female college students is approximately normally distributed with mean m of 67 inches and standard deviation s of 3.95 inches. Show all work.
(A) Find the proportion of female college students whose height is greater than 63 inches.
(B) Find the proportion of female college students whose height is no more than 63 inches.
4. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.45 inches and a standard deviation of 0.45 inches. Show all work.
(A) What percentage of the grapefruits in this orchard have diameters less than 7.1 inches?
(B) What percentage of the grapefruits in this orchard are larger than 6.8 inches?
5. Find the normal approximation for the binomial probability that x = 4, where n = 13 and p = 0.3. Compare this probability to the value of P(x=5) found in Table 2 of Appendix B in your textbook.
6. A set of data is normally distributed with a mean of 100 and standard deviation of 15. · What would be the standard score for a score of 70? · What percentage of scores is between 100 and 70? · What would be the percentile rank for a score of 70?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. A set of 50 data values has a mean of 28 and a variance of 4.
a. Find the standard score (z) for a data value = 31.
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z(31)= (31-28)/4 = 3/4
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b. Find the probability of a data value > 31.
P(x> 31) = P(z > 3/4) = 0.2266
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2.Find the area under the standard normal curve:
a. to the right of z = 2.37
P(z > 2.37) = 0.0089
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b. to the left of z = 2.37
P( z < 2.37) = 1 -0.0089 = 0.9911
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3. Assume that the population of heights of female college students is approximately normally distributed with mean m of 67 inches and standard deviation s of 3.95 inches. Show all work.
(A) Find the proportion of female college students whose height is greater than 63 inches.
z(63) = (63-67)/3.95 = -1.0127
P(x > 63) = P(z > -1.0127) = 0.8444
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(B) Find the proportion of female college students whose height is no more than 63 inches.
P(x <= 63) = 1-0.8444 = 0.1556
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4. The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.45 inches and a standard deviation of 0.45 inches. Show all work.
(A) What percentage of the grapefruits in this orchard have diameters less than 7.1 inches?
Find the z-score of 7.1.
Find the probability z is less than that z-score
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(B) What percentage of the grapefruits in this orchard are larger than 6.8 inches?
Follow the same procedure as in "A".
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5. Find the normal approximation for the binomial probability that x = 4, where n = 13 and p = 0.3. Compare this probability to the value of P(x=5) found in Table 2 of Appendix B in your textbook.
mean = np = 13*0.3 = 3.9
s = sqrt(npq) = sqrt3.9*0.7) = 1.65
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P( x = 4) = P(3.5 < x < 4.5)
z(3.5) = (3.5-3.9)/1.65 = -0.2424
z(4.5) = (4.5-3.9)/1.65 = 0.3636
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P(x = 4) ~ P(-0.2424 < z < 0.3636) = 0.2377
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Check your textbook for P(x = 4) = 0.2337
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6. A set of data is normally distributed with a mean of 100 and standard deviation of 15.
What would be the standard score for a score of 70?
Find the z-score.
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What percentage of scores is between 100 and 70?
Find the z-scores and the percentage between them.
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What would be the percentile rank for a score of 70?
Find the z-score of 70; then find the percentage less than that z-score.
Maybe round to a whole-number value.
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Cheers,
Stan H.
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