Answer:
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 80, \sigma = 4[/tex]
The percentage of cockroaches weighing between 77 grams and 83 grams
This is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 83. So
X = 83
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{83 - 80}{4}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a pvalue of 0.7734
X = 77
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{77 - 80}{4}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] has a pvalue of 0.2266
0.7734 - 0.2266 = 0.5468
Rounded to the nearest whole number, 55%
The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.