Six random samples were taken each year for three years. The table shows the inferred population for each type of fish for each
2 answers:
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Answer:
a) Between 39 and 40 rods can be expected to have a strength less than 39.4 kpsi.
b) 260 rods are expected to have a strength between 39.4 and 60 kpsi
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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:
(a) Assuming a normal distribution, how many rods can be expected to have a strength less than 39.4 kpsi?
The percentage of rods with a stength less than 39.4 is the pvalue of Z when X = 39.4. So
has a pvalue of 0.1314
13.14% of rods have a strength less than 39.4 kpsi.
Out of 300
0.1314*300 = 39.42
Between 39 and 40 rods can be expected to have a strength less than 39.4 kpsi.
(b) How many are expected to have a strength between 39.4 and 60 kpsi?
The percentage of rods with a stength in this interval is the pvalue of Z when X = 60 subtracted by the pvalue of Z when X = 39.4. So
X = 60
has a pvalue of 0.9987
X = 39.4
has a pvalue of 0.1314
0.9987 - 0.1314 = 0.8673
86.73% of the rods are expected to have a strength between 39.4 and 60 kpsi
Out of 300
0.8673*300 = 260
260 rods are expected to have a strength between 39.4 and 60 kpsi