The average time taken to complete an exam, X, follows a normal probability distribution with mean = 60 minutes and standard dev
1 answer:
Answer: b. 0.8413
Step-by-step explanation:
Given : The average time taken to complete an exam, X, follows a normal probability distribution with and
.
Then, the probability that a randomly chosen student will take more than 30 minutes to complete the exam will be :-
[using z-value table]
Hence, the probability that a randomly chosen student will take more than 30 minutes to complete the exam = 0.8413
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Answer:
(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716
(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814
Step-by-step explanation:
We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;
Mean, = 30.05 inches and Standard deviation,
= 0.2 inches
Let X = A sheet selected at random from the population
Here, the standard normal formula is ;
Z = ~ N(0,1)
(a) <em>The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65) </em>
P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)
P(X < 30.65) = P( <
) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425
= 0.9985
P(X <= 30.25) = P( <=
) = P(Z <= 1) = 0.84134
Therefore, P(30.25 < X < 30.65) = 0.9985 - 0.84134 = 0.15716 .
(b)<em> Let Y = Standard Normal Variable is given by N(0,1) </em>
<em> Which means mean of Y = 0 and standard deviation of Y = 1</em>
So, Probability that a standard normal random variable will be between 0.3 and 3.2 = P(0.3 < Y < 3.2) = P(Y < 3.2) - P(Y <= 0.3)
P(Y < 3.2) = P( <
) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688
= 0.99931
P(Y <= 0.3) = P( <=
) = P(Z <= 0.3) = 0.61791
Therefore, P(0.3 < Y < 3.2) = 0.99931 - 0.61791 = 0.3814 .