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Using the normal distribution, we have that:
a) The probability of a student scoring less than 19 is 0.2709 = 27.09%.
b) The probability of a student scoring between 17.7 and 27.1 is 0.6424 = 64.24%.
c) The probability of a student scoring more than 33.1 is 0.0179 = 1.79%.
d) The correct option is: B. The event in part (c) is unusual because its probability is less than 0.05.
<h3>Normal Probability Distribution</h3>The z-score of a measure X of a normally distributed variable with mean and standard deviation
is given by:
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
Item a:
The probability is the <u>p-value of Z when X = 19</u>, hence:
Z = -0.67.
Z = -0.67 has a p-value of 0.2709.
The probability of a student scoring less than 19 is 0.2709 = 27.09%.
Item b:
The probability is the <u>p-value of Z when X = 27.1 subtracted by the p-value of Z when X = 17.7</u>, hence:
X = 27.1:
Z = 0.92.
Z = 0.92 has a p-value of 0.8212.
X = 17.7:
Z = -0.92.
Z = -0.92 has a p-value of 0.1788.
0.8212 - 0.1788 = 0.6424.
The probability of a student scoring between 17.7 and 27.1 is 0.6424 = 64.24%.
Item c:
The probability is <u>one subtracted by the p-value of Z when X = 33.1</u>, hence:
Z = 2.1.
Z = 2.1 has a p-value of 0.9821.
1 - 0.9821 = 0.0179.
The probability of a student scoring more than 33.1 is 0.0179 = 1.79%.
Item d:
Probabilities less than 0.05 are unusual, hence the correct option is:
B. The event in part (c) is unusual because its probability is less than 0.05.
More can be learned about the normal distribution at brainly.com/question/28135235
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