Question

A livestock company reports that the mean weight of a group of young steers is 1088 pounds with a standard deviation of 63 pounds. Based on the model ​N(1088​,63​) for the weights of​ steers, what percent of steers weigh ​a) over 900 ​pounds? ​b) under 1050 ​pounds? ​c) between 1200 and 1250 ​pounds? ​a) 99.86​% of steers have weights above 900 pounds. ​(Round to one decimal place as​ needed.) ​b) 27.43​% of steers have weights below 1050 pounds. ​(Round to one decimal place as​ needed.) ​c) 3.3​% of the steers weigh between 1200 and 1250 pounds. ​(Round to one decimal place as​ needed.)

Can you please show me how to solve this using a TI-84 calculator.

Answer 1

We can solve this problem easily with the help of Excel.

Use the function =NORMDIST(x,mean,standard deviation,cumulative)

where x = hypothesized value

mean = given sample mean

standard deviation = given sample standard deviation

cumulative = TRUE

We are given with:

Mean = 1088 pounds

Standard deviation = 63 pounds

a) over 900 ​pounds?

x = 900

Using the function =NORMDIST(x,mean,standard deviation,cumulative), we get:

=NORMDIST(900,1088,63,TRUE)

This value will give us the region below 900. To get the value over 900, subtract this value from 1.

1 - 0.001422

= 0.998578 = 99.86%

​b) under 1050 ​pounds?

Using the function =NORMDIST(x,mean,standard deviation,cumulative), we get:

=NORMDIST(1050,1088,63,TRUE)

0.274336 = 27.43%

​c) between 1200 and 1250 ​pounds?

As we have to calculate the value in between, we will subtract the higher probability from the lower probability.

Using the function =NORMDIST(x,mean,standard deviation,cumulative), we get:

=NORMDIST(1250,1088,63,TRUE)

0.994936

=NORMDIST(1200,1088,63,TRUE)

0.96228

Required probability is:

0.994936 - 0.96228 = 0.032656 = 3.3%