4 Challenging Statistics Problems! Answer ONLY if you know it 1) Castile soap is
ID: 3132654 • Letter: 4
Question
4 Challenging Statistics Problems! Answer ONLY if you know it
1) Castile soap is favored by some vegetarians, because it contains no animal fats. With the data they've collected, a soap manufacturer has found that 37% of their castile soap bars contain less than or equal to .75 ounces of coconut oil, and 42% of their castile soap bars contain at least 1 ounce of coconut oil. They know from experience that the amount of coconut oil in their soap bars is normally distributed.
Based on this information, which of the following best estimates the mean and standard deviation of the amount of coconut oil in their castile soap bars?
m = 0.91; s = 0.47
m = 1.12; s = 0.58
m = 0.25; s = 0.82
m = 0.84; s = 0.78
m = 0.37; s = 0.66
2) Assume a sample has been selected at random from a large population. Which of the following five statements is correct?
The standard error equals the standard deviation of the sample mean.
The sample standard deviation decreases by one if the value one is subtracted from each sample unit.
If each sample unit is multiplied by the number 2, then the sample standard deviation will increase by the square-root of 2.
If all sample units are equal to each other, then the standard error will be the number 1.
The standard error increases as the sample size increases.
3) A golfer is approaching the fourteenth hole, and he has 2 par-three holes and 3 par-four holes remaining in the match. He knows from experience that he stands a 60% chance of making par or better on each of the par-three holes and a 75% chance of making par or better on each of the par-four hole. What's the probability that he'll shoot par or better on at least four of the last five holes?
0.25
0.69625
0.50625
0.151875
0.30375
4) Rod has taken his bar exam to begin practicing as an attorney. He was informed that he placed in the 86th percentile. Scores for the bar exam were normally distributed with a mean of 177 and a standard deviation of 38. What score did he receive?
199
218
233
246
254
m = 0.91; s = 0.47
m = 1.12; s = 0.58
m = 0.25; s = 0.82
m = 0.84; s = 0.78
m = 0.37; s = 0.66
Explanation / Answer
1) this question will be solved by the z score
Za = (0.75 m) / s
Zb = (1 m) / s
P(Z < Za) = 0.37
therefore the Zscore = 0.33 1/3 as given that 37% contain less than or equal to 0.75 so the z score comes out to be -1/3
P(Z > Zb) = 0.42 as the 42% are containing at least 1 ounce of cocunut , this can also be written as
P(Z < z) = 0.58
therefore the z score = z = 0.20 = 1/5
now we have got two equations
(0.75 m) / s = 1/3 ------A
(1 m) / s = 1/5 -------B
Solving simultaneous equations, BY ELIMINATION METHOD we get:
WE WILL GET M = 0.91
AND S = 0.47
HENCE OPTION A IS CORRECT
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