Question 1185487: 4. Primary care data suggest that many people develop a common cold approximately twice per year. Assuming that: (a) the surveillance year is the same as the calendar year; (b) the risk of developing a cold is equal throughout the year; and (c) the time between colds is normally distributed with a mean of 160 and standard deviation of 40 days, what is the probability of:
(a) Having had a cold in the 4 months preceding a new cold?
(b) Getting another cold within 6 weeks of a previous cold?
(c) Going 200 days or more between colds?
(d) Assume you caught a cold at the beginning of the Autumn Trimester (new country, new friends, new exposures, etc). Based on the data above how likely are you to get another cold by the end of April?
NOTE: For this question, you may assume the following:
(a) a month is 30 days; and
(b)’beginning of Autumn Trimester’ means the beginning of September
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this cold probability problem:
**(a) Probability of a cold in the 4 months preceding a new cold:**
1. **Convert to days:** 4 months * 30 days/month = 120 days
2. **Calculate the z-score:** This tells us how many standard deviations 120 days is away from the mean of 160 days.
z = (x - μ) / σ
z = (120 - 160) / 40
z = -1
3. **Find the probability:** We want the probability of the time between colds being *less than* 120 days. This is equivalent to finding the area to the *left* of z = -1 on the standard normal distribution. Using a z-table or calculator, P(Z < -1) ≈ 0.1587.
**(b) Probability of another cold within 6 weeks of a previous cold:**
1. **Convert to days:** 6 weeks * 7 days/week = 42 days
2. **Calculate the z-score:**
z = (42 - 160) / 40
z = -2.95
3. **Find the probability:** We want the probability of the time between colds being *less than* 42 days. This is the area to the *left* of z = -2.95. Using a z-table or calculator, P(Z < -2.95) ≈ 0.0016.
**(c) Probability of going 200 days or more between colds:**
1. **Calculate the z-score:**
z = (200 - 160) / 40
z = 1
2. **Find the probability:** We want the probability of the time between colds being *greater than or equal to* 200 days. This is the area to the *right* of z = 1. P(Z ≥ 1) = 1 - P(Z < 1). From a z-table, P(Z < 1) ≈ 0.8413. So, P(Z ≥ 1) = 1 - 0.8413 ≈ 0.1587.
**(d) Probability of another cold by the end of April:**
1. **Time elapsed:** From the beginning of September to the end of April is 8 months.
2. **Convert to days:** 8 months * 30 days/month = 240 days.
3. **We want to know the probability of getting a cold within 240 days of the last one.** Since the average time between colds is 160 days, and it's already been 240 days, this is almost certain. Let's calculate the z-score just to be sure.
z = (240-160)/40 = 2
P(Z<2) ≈ 0.9772. So there is a 97.72% chance that you'll get another cold by the end of April.
**In summary:**
* (a) ≈ 0.1587
* (b) ≈ 0.0016
* (c) ≈ 0.1587
* (d) ≈ 0.9772
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