Question

A) The amount of tea leaves in a can from a particular production line is normally distributed with μ = 110 grams and σ = 25 grams. What is the probability that a randomly selected can will contain between 82 and 100 grams of tea leaves?

B) A physical fitness association is including the mile run in its secondary-school fitness test. The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 450 seconds and a standard deviation of 60 seconds. The fitness association wants to recognize the boys whose times are among the top (or fastest) 10% with certificates of recognition. What time would the boys need to beat in order to earn a certificate of recognition from the fitness association? Round to one decimal place.

Answer 1

Question A

Mean = 110

Standard deviation = 25

We need to calculate the probability that a randomly selected can will contain between 82 to 100 grams of tea leaves.

Let X be the random variable indicating the amount of tea leaves in the can. So, we get

Here, z represents the standard normal variable whose mean = 0 and standard deviation = 1. We convert X into the standard normal variable and find the probability values from the table of z.

The probability that a randomly selected can will contain between 82 to 100 grams of tea leaves is 0.26424.

Question B

Mean = 450

Standard deviation = 60

We need to calculate the time which will make the boys among the top 10%

Let X be the random variable indicating the time taken by the boys to complete the mile run. So, we get

putting this in terms of standard normal variable z,

From the table for z, we get

From this, we can calculate the value of X₁ as follows

The boys need to beat the time of 373.2 secs.