How many standard deviations (of X) below the null value is x 5 72.3? b. If x 5 72.3, what is the conclusion using a 5 .002? c. For the test procedure with a 5 .002, what is b(70)? d. If the test procedure with a 5 .002 is used, what n is necessary to ensure that b(70) 5 .01? e. If a level .01 test is used with n 5 100, what is the probability of a type I error when m 5 76?

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b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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