In a survey, 24 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $31 and standard deviation of $6. Construct a confidence interval at a 80% confidence level.

29.4≤μ≤32.6

Step-by-step explanation:

The datas given from the questions are as shown:

Number of people n = 24

Mean xbar= $31

Standard deviation σ = $6

Confidence Interval formula is expressed as:

CI = xbar ± Z(σ/√n)

Z value for 80% confidence interval is 1.282

Substituting the values into the Confidence Interval formula will give;

CI = 31 ± 1.282{6/√24}

CI = 31 ± 1.282(1.225)

CI = 31 ± 1.57045

CI = 31+1.57045 and 31-1.57045

CI = (29.42955, 32.57045)

CI = (29.4, 32.6) to 1dp

The confidence interval will be within the range 29.4≤μ≤32.6

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b