The manager of a computer shop is recording the time taken for customers to decide of which computer and accessories they buy from the time they enter the store. From the previous data, it is known that the average ‘decision’ time was 45 minutes. The manager assumes a normally distributed population with a standard deviation of 10 minutesa) What is the probability that a customer will take more than 60 minutes?b) What proportion of customers will take between 35 and 60 minutes?c) From a sample of 36 customers, what is the probability that the sample mean will be less than 40 minutes?d) Is the sample taken in c) assumed to be normally distributed? Why? What theorem is used to support this assumption?
Question
The manager of a computer shop is recording the time taken for customers to decide of which computer and accessories they buy from the time they enter the store. From the previous data, it is known that the average ‘decision’ time was 45 minutes. The manager assumes a normally distributed population with a standard deviation of 10 minutesa) What is the probability that a customer will take more than 60 minutes?b) What proportion of customers will take between 35 and 60 minutes?c) From a sample of 36 customers, what is the probability that the sample mean will be less than 40 minutes?d) Is the sample taken in c) assumed to be normally distributed? Why? What theorem is used to support this assumption?
Solution
a) To find the probability that a customer will take more than 60 minutes, we first need to calculate the z-score. The z-score is calculated as (X - μ) / σ, where X is the value we're interested in, μ is the mean, and σ is the standard deviation. So, the z-score for 60 minutes is (60 - 45) / 10 = 1.5. Looking up this z-score in a standard normal distribution table, we find that the probability of a value being less than 60 minutes is approximately 0.9332. Therefore, the probability of a value being more than 60 minutes is 1 - 0.9332 = 0.0668, or 6.68%.
b) To find the proportion of customers who will take between 35 and 60 minutes, we need to calculate the z-scores for both 35 and 60 minutes. The z-score for 35 minutes is (35 - 45) / 10 = -1. The z-score for 60 minutes is 1.5, as we calculated earlier. Looking up these z-scores in a standard normal distribution table, we find that the probability of a value being less than 35 minutes is approximately 0.1587, and the probability of a value being less than 60 minutes is approximately 0.9332. Therefore, the proportion of customers who will take between 35 and 60 minutes is 0.9332 - 0.1587 = 0.7745, or 77.45%.
c) To find the probability that the sample mean will be less than 40 minutes from a sample of 36 customers, we first need to calculate the standard error, which is σ / √n, where σ is the standard deviation and n is the sample size. So, the standard error is 10 / √36 = 1.67. Then, we calculate the z-score for 40 minutes, which is (40 - 45) / 1.67 = -3. The probability of a value being less than 40 minutes is approximately 0.0013, or 0.13%, according to the standard normal distribution table.
d) Yes, the sample taken in c) is assumed to be normally distributed. This is because of the Central Limit Theorem, which states that the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution, provided the sample size is sufficiently large (usually n > 30).
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