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2 years ago

A baseball player has a batting average of 0.235. What is the probability that he has exactly 2 hits in his next 7

Mathematics

1 answer:

Ksivusya [100]2 years ago

8 0

Probabilities are used to determine the chances of an event.

The probability that he hits exactly 2 in the next 7 bats is 0.3039

The given parameters are:

$\mathbf{p = 0.235}$ --- probability that he hits a bat

The question is an illustration of binomial probability, and it is represented as:

$\mathbf{Pr = ^nC_xp^x(1 - p)^{n -x}}$

In this case:

$\mathbf{n = 7}$ --- number of bats

$\mathbf{x = 2}$ --- number of hits

So, we have:

$\mathbf{Pr = ^nC_xp^xq^{n -x}}$

$\mathbf{Pr = ^7C_2 \times (0.235)^2 \times (1 - 0.235)^{7 -2}}$

$\mathbf{Pr = 21 \times (0.235)^2 \times (0.765)^5}$

$\mathbf{Pr = 0.3039}$

Hence, the probability that he hits exactly 2 in the next 7 bats is 0.3039

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Step-by-step explanation:

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2 years ago

A group of 121 Internet users 50-65 years of age were randomly chosen and asked to report the weekly number of hours they spend

Luda [366]

Answer: We support the null hypothesis.

Step-by-step explanation:

Given : The purpose of the study was to determine whether the mean weekly number of hours that Internet users in that age group spend online differs from the mean for Internet users in general which is 12.50 hours.

The set of hypothesis for the given claim :-

$H_0:\mu=12.50\\\\ H_a:\mu\neq12.50$

A 95% confidence interval for the mean weekly number of hours that Internet users in that age group spend online was calculated to be (11.439 , 12.577).

Since 12.50 contained in the 95% confidence interval (11.439 , 12.577).

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So the final conclusion will be, we have evidence to support the claim that Internet users in that age group spend online differs from the mean for Internet users in general which is 12.50 hours.

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Step-by-step explanation:

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3 years ago

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