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3 years ago

Can someone help me please? ill give brainliest. please help this is due soon :(

Mathematics

1 answer:

MAVERICK [17]3 years ago

4 0

Answer:

B & C

Step-by-step explanation:

Okay I don't know much you need explained so I'm gonna explain pretty much everything. Starting off, you need to follow the PEMDAS rules. PEMDAS stands for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction. So when solving you have to do it in this order.

A-(5)(0.6)*7=21

B-(4/5)(15)+8=(0.8)(15)+8=20

C-4(8+2)/2=4(10)/2=40/2=20

D-14+4(2+3)=14+4(5)=14+20=34

If you need any clarification just lmk :)

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b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

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d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538$

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{36 - 38.4}{0.8538}$

$Z = -2.81$

$Z = -2.81$ has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{41 - 38.4}{0.8538}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

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Now $n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08$

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

$Z = \frac{X - \mu}{s}$

$Z = \frac{36 - 38.4}{1.08}$