If it is 3;50 and i have to leave at 4;45 how long do i have

Mathematics

1 answer:

MArishka [77]3 years ago

6 0

Answer: 55. minutes

Step-by-step explanation:

10+45

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Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

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a) 0.5588

b) 0.9984

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24

Standard Deviation, σ = 6.4

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(score between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{6.4} \leq z \leq \displaystyle\frac{30-24}{6.4}) = P(-0.62 \leq z \leq 0.94)\\\\= P(z \leq 0.94) - P(z < -0.62)\\= 0.8264 - 0.2676 = 0.5588 = 55.88\%$

$P(20 \leq x \leq 30) = 55.88\%$

b) Sampling distribution

Sample size, n = 22

The sample will follow a normal distribution with mean 24 and standard deviation,

$s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{6.4}{\sqrt{22}} =1.36$

c) P(mean score of sample is between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{1.36} \leq z \leq \displaystyle\frac{30-24}{1.36}) = P(-2.94 \leq z \leq 4.41)\\\\= P(z \leq 4.41) - P(z < -2.94)\\= 1 - 0.0016 = 0.9984 = 99.84\%$