*The width of semiconductor is assumed to be normally distributed with a mean of
ID: 2921701 • Letter: #
Question
*The width of semiconductor is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer.
a. What is the probability that the width is between 0.47 and 0.63 micrometer?
b. Below what value is the width of 95% of semiconductor?
*Suppose that 80% of all STAT210 students like MacBook Air. What is the probability that of 20 randomly chosen STAT210 students,
a. Ten students like MacBook Air?
b. Between 7 and 10 (inclusive) students like MacBook Air?
c. Calculate the mean and standard of this distribution
Explanation / Answer
mean = 0.5 , sd= 0.05
a)
Z = (X * 0.5 )/0.05
P( 0.47 <X<0.63)
= P ( -0.6 < Z< 2.6)
= 0.721
b)
z* = 1.96 for 95 %
hence confidence interval is
(0.5 -1.96*0.05),)0.5+1.96*0.05)
=(0.402, 0.598)
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