Question
The amount of Gasoline sold daily at M&M Gas Station is uniformly distributed with a minimum of 2,000 gallons and a maximum of 5, 000
The amount of Gasoline sold daily at M&M Gas Station is uniformly distributed with a minimum of 2,000 gallons and a maximum of 5, 000 gallons.
- 1. (a) What is probability that sales will fall between 2,500 and 3,000 gallons? A) 0.1667 B) 0.1766 C) 0.1787 D) 0.7167
- 1. (b) What is the probability that M&M will sell at least 4,000 gallons? A) 0.3333 B) -0.3333 C) -0.6666 D) 0.1667
- 1. ( c) What is the probability that the station will sell exactly 3,000 gallons? A) 0.001 B) 0.000 C) 1.000 D) 0.999
The amount of time it takes a student to complete Statistics quiz is uniformly distributed with a minimum of 30 minutes and a maximum of 60 minutes. One student is selected at random. Find the probability of the following events:
- 2(a) What is the probability that the student requires more than 55 minutes to complete the quiz? A) 0.1667 B) 0.7661 C) 0.6761 D) 0.6667
- 2(b) What is the probability that the student completes the quiz in a time between 30 and 40 minutes? A) 0.3333 B) 0.6666 C) 0.1667 D) 0.6716
- 2( c) What is the probability that the student completes the quiz in exactly 36.69 minutes ? A) -0.1667 B) -0.7661 C) 0.0000 D) 0.0001
3. Suppose that at another gas station the daily demand for regular gasoline is normally distributed with a mean of 1,000 gallons and a standard deviation of 100 gallons. The manager has just opened the gas station for business and notes that there is exactly 1,100 gallons of regular gasoline in storage. The next delivery is scheduled later today at the close of business . The manager would like to know the probability that he will have enough regular gasoline to satisfy today's demands . Note: The amount of gasoline on hand will be sufficient to satisfy the demand if the demand is less than the supply.
- 3(a) What is P(Z <1.00) ? A) 0.8413 B) 0.8443 C) 0.1587 D) 0.8643
- 3(b) What is P(-0.71< Z < 0.92) ? A) 0.5823 B) 0.5283 C) 0.8212 D) 0.2389
- 3(c)What is P(Z > 3.09) ? A) 0.0010 B) 0.0110 C) 0.1010 D) 0.1110
4. The following problem will demonstrate why risk is measured by the variance and the standard deviation. Consider an investment whose return is normally distributed with a mean of 10% and a standard deviation of 5% . We must determine the probability of losing money; and the probability of losing money when the standard deviation doubles i.e. = 10%. The investment loses money when the return is negative. That is, we need to determine P(X < 0).
- 4(a) What is the value of P(X < 0) when = 5% ? A) 0.2280 B) 0.0228 C) 0.0828 D) 0.8228
- 4(b) What is the value of P(X < 0) when =10% ? A) 0.0158 B) 0.1587 C) 0.8517 D) 0.5817
5(a) Find Z.05 A) 1.546 B) 1.645 C) 1.960 D) 1.665
5(b) Find Z.025 A) 1.690 B) 1.960 C) 2.960 D) 1.666
5( c) Find Z.045 A) 1.70 B) 1.86 C) 1.79 D) 1.98
5(d) Find Z.02 A) 2.05 B) 2.33 C) 2.44 D) 2.96
6. The probability distribution of of the random variable X - throwing a fair die infinitely many times is shown in the table below: Use the following table to answer the following:
X 1 2 3 4 5 6
P(x) 1/ 6 1/ 6 1 /6 1 /6 1 /6 1/ 6
- 6(a) What is E(x), the population mean ? A) 5.3 B) 3.5 C) 6.5 D) 6.3
- 6(b) What is the Variance ? A) 2.29 B) 2.92 C) 1.46 D) 1.64
- 6( c) What is the standard deviation ? A) 1.17 B) 1.71 C) 2.17 D) 2.71
8.A sample of n = 16 observations is drawn from a normal population/distribution with =1,000 and = 200. Find the following:
- 8(a) P(x>1,050). A) 0.1587 B) 0.8157 C) 0.5187 D) 0.8751
- 8(b) P(x< 960) A) 0.2119 B) 0.2991 C) 0.2009 D) 0.2900
10. A medical statistician wants to estimate the average weight loss of people who are on a new diet plan. In a preliminary study, he guesses that the standard deviation of the population of weight losses is about 10 pounds. How large a sample should he take to estimate the mean weight loss to within 2 pounds, with a 90% confidence? 10(a) : A) 68 B) 86 C) 96 D) 130
10(b): for confidence of 98% A) 136 B) 183 C) 213 D) 231
11(a) Calculate the LCL(lower confidence limit) ( forx = 370.16, confidence = 95%, n = 25 A) 340.76 B) 304.76 C) 376.04 D) 346.07
11(b) Calculate the UCL( upper confidence limit) for x = 370.16, confidence = 95%, n = 25 A) 399.65 B) 399.56 C) 309.56 D) 509.63
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