a statistics instructor collected data on the time it takes the students to complete a test. the test taking time is uniformly distributed within a range of 35 minutes to 55 minutes. determine the standard deviation,

what is the probability a particular student will take less than 45 minutes
what is the probability a particular student will take between 45 and 50 minutes

To determine the standard deviation, we can use the formula for the standard deviation of a Uniform distribution given its range.

The formula for the standard deviation (σ) of a Uniform distribution is:

σ = (b - a) / √12

Where:
- a is the lower limit of the range (35 minutes in this case)
- b is the upper limit of the range (55 minutes in this case)

Substituting the values into the formula:

σ = (55 - 35) / √12
= 20 / √12
≈ 5.77 minutes

Therefore, the standard deviation of the test taking time is approximately 5.77 minutes.

To find the probability that a particular student will take less than 45 minutes, we need to standardize the value using the z-score and then use the standard normal distribution table.

The z-score formula is:

z = (x - μ) / σ

Where:
- x is the value we want to find the probability for (45 minutes in this case)
- μ is the mean of the distribution (which is the average of the lower and upper limits, i.e., (35 + 55) / 2 = 45 minutes)
- σ is the standard deviation (5.77 minutes)

Substituting the values into the formula:

z = (45 - 45) / 5.77
= 0 / 5.77
= 0

Since the z-score is 0, we can look up the corresponding probability in the standard normal distribution table. The probability of getting a z-score of 0 (or any value between -1 and 1) is 0.5. Therefore, the probability that a particular student will take less than 45 minutes is 0.5 or 50%.

To find the probability that a particular student will take between 45 and 50 minutes, we need to calculate the z-scores for both values (45 and 50) using the formula mentioned earlier.

For 45 minutes:
z1 = (45 - 45) / 5.77
= 0

For 50 minutes:
z2 = (50 - 45) / 5.77
= 5 / 5.77
≈ 0.87

Now, we need to find the probability corresponding to both z1 and z2 from the standard normal distribution table. The table gives the cumulative probability, so we subtract the probability corresponding to z1 from the probability corresponding to z2.

P(45 < x < 50) = P(z1 < z < z2)
= P(0 < z < 0.87)

By referring to the standard normal distribution table, we can find that the probability of having a z-score between 0 and 0.87 is approximately 0.31. Therefore, the probability that a particular student will take between 45 and 50 minutes is 0.31 or 31%.