What is the probability of randomly selecting a z-score greater than z = -0.75 from a normal distribution? %3D O p= 0.4878 O p= 0.2734 Op= 0.7734 Op 0.2266

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Q: The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally…

A: Given that mu =1252, sigma= 129

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A: The z-score is known as the standard score.

Q: The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately…

A: Hey there! Thank you for posting the question. Since your question has more than 3 parts, we are…

Q: for a normal standard distribution, what is the probability that z>=.0.04?

A: We have given the normal standard deviation. The given z-score is- z≥0.04 We have to find the…

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A: Given : The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately…

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A: we want to find probability of sampling an individual with a score greater than 48

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Q: The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally…

A: Given Data: Let X represent the random variable   Mean=1252 SD=129   X~ Normal (1252, 1292)

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Q: The entire family of normal distribution is differentiated by Select one: a. none of the above

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A: Given μ= 55 and a σ=5.

Q: e probability of randomly selecting a z-score greater than z = +1.25 from a normal distribution?…

A: You can find the required area by using the standard normal table.  Now,

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Q: The 80th percentile of the standard normal (Z) distribution is 84. O True False

A: Step1: Given distribution is standard normal. i.e, Z~N(0,1) Hence, μ=0 and σ=1

Q: The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately…

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Q: The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally…

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Q: The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally…

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A: Given data,μ=21σ=10P(15<X<25)=?

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A: A standard normal variable z follows normal distribution with mean 0 and standard deviation 1.

Q: The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally…

A: GivenMean(μ)=1252standard deviation(σ)=129

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Q: The number of chocolate chips in an 18-ounce bag of chocolate chip cookies is approximately normally…

A: GivenMean(μ)=1252standard deviation(σ)=129

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A: Given that, Z score = -0.25 (Left tailed)

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A: probability of randomly selecting a z-score less than z=-1.55 from a normal distribution

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A: As per company guidelines I solved exactly three subparts of the question

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Q: Find Pas, that is find 90 percentile of z-score from the standard normal distribution. 6.

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A: As per our guidelines we are suppose to answer three sub parts. Given,mean(μ)=1252standard…

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A: For standard normal distribution, μ=0 and σ=1 90% percentile means, PZ≤90th percentile value=90%

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