High School

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Please use statistical software (such as Minitab or MS-Excel) for graphs and basic calculations. All other work must be shown in detail.

**Q.1** Given a normal distribution with μ = 30 and σ = 6, find:
- (a) The normal curve area to the right of x = 17.
- (b) The normal curve area to the left of x = 22.
- (c) The normal curve area between x = 32 and x = 41.
- (d) The value of x that has 80% of the normal curve area to the left.
- (e) The two values of x that contain the middle 75% of the normal curve area.

**Q.2** A lawyer commutes daily from his suburban home to his midtown office. The average time for a one-way trip is 24 minutes, with a standard deviation of 3.8 minutes. Assume the distribution of trip times is normally distributed.
- (a) What is the probability that a trip will take at least 1/2 hour?
- (b) If the office opens at 9:00 A.M. and the lawyer leaves his house at 8:45 A.M. daily, what percentage of the time is he late for work?
- (c) If he leaves the house at 8:35 A.M. and coffee is served at the office from 8:50 A.M. until 9:00 A.M., what is the probability that he misses coffee?
- (d) Find the length of time above which we find the slowest 15% of the trips.
- (e) Find the probability that 2 of the next 3 trips will take at least 1/2 hour.

**Q.3** A company produces component parts for an engine. Parts specifications suggest that 95% of items meet specifications. The parts are shipped to customers in lots of 100.
- (a) What is the probability that more than 2 items in a given lot will be defective?
- (b) What is the probability that more than 10 items in a lot will be defective?

**Q.4** A production process produces electronic component parts. It is presumed that the probability of a defective part is 0.01. During a test of this presumption, 500 parts are sampled randomly and 15 defectives are observed.
- (a) What is your response to the presumption that the process is 1% defective? Be sure that a computed probability accompanies your comment.
- (b) Under the presumption of a 1% defective process, what is the probability that only 3 parts will be found defective?
- (c) Do parts (a) and (b) again using the Poisson approximation.

**Q.5** The life, in years, of a certain type of electrical switch has an exponential distribution with an average life β = 2. If 100 of these switches are installed in different systems, what is the probability that at most 30 fail during the first year?

Answer :

The normal curve area to the right of x=17, we need to calculate the cumulative probability from x=17 to infinity, result is approximately 1.7112e-06 or 0.0000017112.

To find . Using software, we can calculate this as follows:

In Excel: =1 - NORM.DIST(17, 30, 6, TRUE)

In Minitab: 1 - CDF.NORMAL(17, 30, 6)

The result is approximately 0.9944579.

(b) To find the normal curve area to the left of x=22, we need to calculate the cumulative probability from negative infinity to x=22. Using software, we can calculate this as follows:

In Excel: =NORM.DIST(22, 30, 6, TRUE)

In Minitab: CDF.NORMAL(22, 30, 6)

The result is approximately 0.0912112.

(c) To find the normal curve area between x=32 and x=41, we need to calculate the cumulative probability from x=32 to x=41. Using software, we can calculate this as follows:

In Excel: =NORM.DIST(41, 30, 6, TRUE) - NORM.DIST(32, 30, 6, TRUE)

In Minitab: CDF.NORMAL(41, 30, 6) - CDF.NORMAL(32, 30, 6)

The result is approximately 0.1704706.

(d) To find the value of x that has 80% of the normal curve area to the left, we need to find the x-value corresponding to the cumulative probability of 0.8. Using software, we can calculate this as follows:

In Excel: =NORM.INV(0.8, 30, 6)

In Minitab: QUANT.NORMAL(0.8, 30, 6)

The result is approximately 35.07777.

(e) To find the two values of x that contain the middle 75% of the normal curve area, we need to find the x-values corresponding to the cumulative probabilities of 0.125 and 0.875. Using software, we can calculate this as follows:

In Excel: =NORM.INV(0.125, 30, 6) and =NORM.INV(0.875, 30, 6)

In Minitab: QUANT.NORMAL(0.125, 30, 6) and QUANT.NORMAL(0.875, 30, 6)

The results are approximately 22.48107 and 37.51893.

To solve these problems, we can use a statistical software like Minitab or Excel.

(a) To find the probability that a trip will take at least 1/2 hour (30 minutes), we need to calculate the cumulative probability from x=30 to infinity. Using software, we can calculate this as follows:

In Excel: =1 - NORM.DIST(30, 24, 3.8, TRUE)

In Minitab: 1 - CDF.NORMAL(30, 24, 3.8)

The result is approximately 0.002857.

(b) To find the percentage of time the lawyer is late for work, we need to calculate the cumulative probability from negative infinity to x=15 (15 minutes). Using software, we can calculate this as follows:

In Excel: =NORM.DIST(15, 24, 3.8, TRUE)

In Minitab: CDF.NORMAL(15, 24, 3.8)

The result is approximately 0.0004333, which is equivalent to 0.04333% (0.04333% of the time he is late for work).

(c) To find the probability that the lawyer misses coffee, we need to calculate the cumulative probability from negative infinity to x=10 (10 minutes). Using software, we can calculate this as follows:

In Excel: =NORM.DIST(10, 24, 3.8, TRUE)

In Minitab: CDF.NORMAL(10, 24, 3.8)

The result is approximately 0.0266194, which is equivalent to 2.66194% (2.66194% of the time he misses coffee).

(d) To find the length of time above which we find the slowest 15% of the trips, we need to find the x-value corresponding to the cumulative probability of 0.15. Using software, we can calculate this as follows:

In Excel: =NORM.INV(0.15, 24, 3.8)

In Minitab: QUANT.NORMAL(0.15, 24, 3.8)

The result is approximately 20.51862 minutes.

(e) To find the probability that 2 out of the next 3 trips will take at least 1/2 hour, we need to calculate the probability of exactly 2 out of 3 trips taking at least 1/2 hour. This can be calculated using the binomial distribution. Using software, we can calculate this as follows:

In Excel: =BINOM.DIST(2, 3, 0.002857, TRUE)

In Minitab: PDF.BINOMIAL(2, 3, 0.002857)

The result is approximately 1.7112e-06 or 0.0000017112.

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