Question 232356
Total Number of Peas (n) = 428+152= 580
Sample proportion of yellow peas (p) = 152/580 = 0.262
And q = 1-p = 1-0.262 = 0.738

Here the sample size (580) is greater than 30, using Central limit theorem we can expect it approximately follows normal distribution.

95% confidence interval estimate of the percentage of yellow peas = p (plus or minus)zat0.95*sqrt(p*q/n)
Standard Normal score at 5% level of significance  = 1.96 
95% confidence interval estimate of the percentage of yellow peas is
=  0.262(plus or minus)1.96*sqrt(0.262*0.738/580)
= 0.262(plus or minus)0.0183
= (0.2437, 0.2803)