Suppose Z follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of c so that the following is true. P(-0.88≤Z≤c)=0.7909 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places. 5

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A: The information provided in the question are as follows : Population mean (µ) = 6.3 Sample size (n)…

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A:

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A: Given that  Mean=μ=36 Standard deviation=?=10 X~N( μ = 36 , ? = 10 ) Z-score =( x - μ )/?

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A: given data, normal distribution , μ=100σ=15we have to find out the shaded region for the given…

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A:

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A: We have given information, The given value are discreate. The probability of fewer than 43…

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A: (a)   Determine the distribution of X.   The distribution of X is determined below as follows:   Let…

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A: Population mean, μ=100 Population sd, σ=15

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A: We have given that  X~N( 0 , 1 ) mu=0 , sigma =1 Z-score =(x-mu)/sigma

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A: given data,normal distribution μ=105σ=15probabilty that selected asults has an IDQ score less than…

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A: Solution is given below:-

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A: Givens=4n=26c=0.90

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A: We have given that The probability of at least 44 boys

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A: The mean would be midpoint between the ends. μ = (0 + 7 )/2 = 3.5 minutes

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A: Chebyshev's inequality : Chebyshev's inequality then implies that the likelihood of an observation…

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A: It is needed to find the value of  for 40 degrees of freedom.

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A: Given information Mean = 100 Standard deviation = 15 P(X > x) = 0.65 P(X < x) = 1 – 0.65 =…

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A: The given degrees of freedom is (55, 77).

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A: The mean and standard deviation are 100 and 15.

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A: Given:

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A: As per guideline expert have to answer first 3 subparts only.

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A: From the given graph,  n=16 p=0.55   By using the normal approximation method,  X~Nnp,…

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A: According to the given information, we have Mean = 11 Standard deviation = 4 The distribution is…

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A: From the provided information, Mean (µ) = 100.6 Standard deviation (σ) = 16.6 X~N (100.6, 16.6)

Q: The indicated IQ score, x, is (Round to one decimal place as needed.)

A: Let , X→N(μ=100 , σ=15) Our aim is to find the value of the indicated IQ score.

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A: GivenMean(μ)=0standard deviation(σ)=1

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A: given data normal distributionμ = 66σ= 2.5x = height of individual

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A: Solution Z be standard normal variate with mean 0 and sd 1

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A:

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A: n=11

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