Question

Oxnard College reports your GPA on a 4.0 scale. The "4" in the 4.0 GPA scale is equivalent to an A, "3" is equivalent to a B, "2" is equivalent to a C, "1" is equivalent to a D, and "0" is equivalent to a F.

Last semester, a student recieved the following grades

C in a 3 units Math course
F in a 4 units Music course
A in a 4 units Chemistry course
F in a 5 units Journalism course

What is the student's GPA for that term? Round to two decimal places.

Student's GPA =

54 randomly selected students were asked how many siblings were in their family. Let X represent the number of pairs of siblings in the student's family. The results are as follows:

# of Siblings	0	1	2	3	4	5	6
Frequency	6	7	9	6	10	7	9

Round all your answers to 4 decimal places where possible.

The mean is:

The median is:

The sample standard deviation is:

The first quartile is:

The third quartile is:

What percent of the respondents have at least 2 siblings? %

84% of all respondents have fewer than how many siblings?

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 39 ounces and a standard deviation of 6 ounces.

a) 99.7% of the widget weights lie between  and

b) What percentage of the widget weights lie between 27 and 57 ounces?  %

c) What percentage of the widget weights lie above 33 ?  %

Answer 1

3)

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 39 ounces and a standard deviation of 6 ounces.

a) about 99.7% of observation of data lie within 3 std dev away from mean

µ±3σ = (21,57)

99.7% of the widget weights lie between 21 and 57

b)

we need to calculate probability for ,                      
P (   27   < X <   57   )      
=P( (27-39)/6 < (X-µ)/σ < (57-39)/6 )                      
                      
P (    -2.000   < Z <    3.000   ) = (95/2)+(99.7/2) = 97.35%
  

97.35 percentage of the widget weights lie between 27 and 57 ounces

c) P ( X ≥   33   ) = P( (X-µ)/σ ≥ (33-39) / 6)      
= P(Z ≥   -1.00   )

about 68% of observation of data lie within 1 std dev away from mean

(1-(1-0.68)/2)) = 84% percentage of the widget weights lie above 33