Answer to Question #139996 in Statistics and Probability for Manqoba Gebashe

Question #139996
The time (X) needed to complete a final examination in a particular college course is normally distributed with a mean of 61 minutes and a standard deviation of 9 minutes.

1.1) What is the probability of completing the exam in less than 72 minutes (inclusive)? (4)

1.2) What is the probability that a student will complete the exam in more than 75 minutes (inclusive) but less than 80 minutes(inclusive)? (6)
1
Expert's answer
2020-10-25T18:53:52-0400

Let "X=" the time needed to complete a final examination: "X\\sim N(\\mu, \\sigma^2)."

Then "Z=\\dfrac{X-\\mu}{\\sigma}\\sim N(0, 1)"

Given "\\mu=61, \\sigma=9"

1.1


"P(X\\leq72)=P(Z\\leq\\dfrac{72-61}{9})=P(Z\\leq\\dfrac{11}{9})\\approx"

"\\approx0.889188"

1.2


"P(75\\leq X\\leq80)=P(X\\leq80)-P(X<75)="

"=P(Z\\leq\\dfrac{80-61}{9})-P(Z<\\dfrac{75-61}{9})="

"=P(Z\\leq\\dfrac{19}{9})-P(Z<\\dfrac{14}{9})\\approx"

"\\approx0.9826186-0.9400931\\approx0.042526"



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Comments

Assignment Expert
02.11.20, 22:37

Dear Kenneth Haindongo, please use the panel for submitting new questions.

Kenneth Haindongo
31.10.20, 18:37

A research firm conducted a survey to determine the mean amount of money smokers spend on cigarettes during a day. A sample of 100 smokers revealed that the sample mean is N$ 5.24 and sample standard deviation is N$ 2.18 Assume that the sample was drawn from a normal population. 2.1 Find the point estimate of the population mean. 2.2 Determine the lower limit of the 95% conference interval for estimating the unknown population mean

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