1. Assume that x has a normal distribution with the specified mean and standard
ID: 3178149 • Letter: 1
Question
1. Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)
= 51; = 14
P(40 x 47) =
2. Find z such that 9.5% of the standard normal curve lies to the right of z. (Round your answer to two decimal places.)
z =
3. Find the z value such that 96% of the standard normal curve lies between z and z. (Round your answer to two decimal places.)
z =
Sketch the area described.
4. Suppose x has a distribution with = 13 and = 10.
(a) If a random sample of size n = 43 is drawn, find x, x and P(13 x 15). (Round x to two decimal places and the probability to four decimal places.)
(b) If a random sample of size n = 64 is drawn, find x, x and P(13 x 15). (Round x to two decimal places and the probability to four decimal places.)
(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is the (same as, larger than, smaller) than part (a) because of the (larger, smaller, same) sample size. Therefore, the distribution about x is (the same, narrower, wider).
Explanation / Answer
1 ans ) The transform Z = (X - mu)/sd is a standar normal
= 51; = 14
Z1=( 40-51) / 14 = - 0.7857
Z2 =( 47-51) /14 = - 0.2857 from standard z table we have values
P(40 x 47) = 0.3897 -0.2266 =0.1631
2 ans ) Z = invNorm(.095)
= - 1.94
3 ans) The proportion lying in the two tails of N(0,1) is 1-0.96=0.04, so proportion in each tail is 0.02.
We thus want the z-value such that the proportion from - to z =1-0.02=0.98
From tables of z-values this is z=0.8365
4 ans ) a)
Mean = 13
Standard deviation = 10
Standard error / n = 10 / 43 = 1.524
standardize xbar to z = (xbar - ) / ( / n )
P( 13 < xbar < 15) = P[( 13 - 13) / 1.524 < z < ( 15 - 13) / 1.524]
P( 0 < z < 1.312) = 0.9049
(from normal probability table)
b)
Mean = 13
Standard deviation = 10
Standard error / n = 10 / 64 = 1.25
standardize xbar to z = (xbar - ) / ( / n )
P( 13 < xbar < 15) = P[( 13 - 13) / 1.25 < z < ( 15 - 13) / 1.25]
P( 0 < z < 1.6) =0.9452
c)
The standard deviation (of the mean) is smaller in (b) ; dividing by a smaller number makes the area larger.
Therefore, the distribution about x is wider.
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