Step by step solutions please and thank you :) If Z is a standard normal variabl

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Step by step solutions please and thank you :) If Z is a standard normal variable, find the probability. The probability that Z lies between 1.10 and -0.36 0.2239 -0.2237 0.2237 0.4951 Assume that X has a normal distribution, and find the indicated probability. The mean is mu = 60.0 and the standard deviation is sigma = 4.0. Find the probability that X is less than 53.0 0.0401 0.9599 0.5589 0.0802 The mean is mu = 15.2 and the standard deviation is sigma= 0.9. Find the probability that X is greater than 17. 0.0228 0.9821 0.9713 0.9772 The mean is mu = 137.0 and the standard deviation is sigma = 5.3. Find the probability that X is between 134.4 and 1401. 0.6242 0.4069 0.8138 1.0311 Solve the problem. Scores on a test are normally distributed with a mean of 65.3 and a standard deviation of 10.3. Find P_81, which separates the bottom 81% from the top 19%. 0.291 74.4 68.3 0.88 Find the indicated probability. The incomes of trainees at a local mill are normally distributed with a mean of $1100 and a standard deviation of %150. What percentage of trainees earn less than $900 a month? 9.18 % 90.82% 40.82% 35.31% Solve the problem. The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 91 inches, and a standard deviation of 14 inches. What is the probability that the mean annual precipitation during 49 randomly picked years will be less than 93.8 inches? 0.5808 0.9192 0.0808 0.4192 Assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 90 women are randomly selected, find the probability that they have a mean height between 62.9 inches and 64.0 inches. 0.0424 0.1739 0.7248 0.9318 A final exam in Math 160 has a mean of 73 with standard deviation 78. If 24 students are randomly selected, find the probability that the mean of theory test scores is greater than 71. 0.8962 0.9012 0.5036 0.0008 Use the normal distribution to approximate the desired probability. Metra reports that 74% of its trains are on time. A check of 60 randomly selected trains shows that 38 of them arrived on time. Find the probability that among the 60 trains, 38 or fewer arrive on time. Based on the result, does it seem plausible that the "on-time" rate of 74% could be correct? .0316, yes .0409, yes .0409, no .0316, no

Explanation / Answer

2)
mean = 60 , sd = 4 , x = 53
z = ( x - mean) / s
= ( 53 - 60) / 4
= -1.75
we need to find P( z < -1.75)
P(X < 53) = P( z < -1.75) = 0.04006

3)
mean = 15.2 , sd = 0.9 , x = 17
z = ( x - mean) / s
= ( 17 - 15.2) / 0.9
= 2
we need to find P( z > 2)
P(X > 17) = P( z >2) = 0.02275

4)
mean = 137 , sd = 5.3 , x = 134.4 and 140.1
z = ( x - mean) / s
= ( 134.4- 137) / 5.3
= -0.490

z = ( x - mean) / s
= ( 140.1- 137) / 5.3
= 0.585

we need to find P( -0.490 < z <0.485)
P(134.4 < X < 140.1) =P( -0.490 < z <0.485) = 0.4088

8)
mean = 63.6 , sd = 2.5 , x = 62.9 and 64 , n = 90
z = ( x - mean) / (s/sqrt(n))
= ( 62.9- 63.6) / (2.5 / sqrt(90))
= -2.656

z = ( x - mean) / (s/sqrt(n))
= (64 - 63.6) / (2.5 / sqrt(90))
= 1.517

we need to find P( -2.656 < z <1.517)
P(62.9 < X < 64) =P( -2.656 < z <1.517) = 0.9315

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