The distribution of the weights of loaves of bread from a certain bakery follows approximately a normal distribution. based on a very large sample, it was found that 10 percent of the loaves weighed less than 15.34 ounces and 20 percent of the loaves weighed more than 16.31 ounces. what are the mean and standard deviation of the distribution of the weights of the loaves of bread?

μ=23.710740....

σ=3.592592....

Step-by-step explanation:

At first, we are told that it´s a normal distribution and the following:

P(x≤15.34)=0.1

P(x≤16.31)=0.2

We start looking for those values (0.1 and 0.2) and the probabilities that gives us those values in a normal standard distribution. There´s some tools that´ll help us out but i´ll be using a chart with the values of some probabilities in a normal standard distribution (That it´s attached).

The probability of P(z≤ a.b) where "a" is the whole part of a number and "b" the decimal part is in the coordinates (a,b). For example for our problem:

P(z≤-2.33)=0.099≈0.1

P(z≤-2.06)=0.197≈0.2

To find a probability of a normal (not standard) distribution we use a method called Normalize that proceeds:

P(x≤a)=P( (x-μ)/σ ≤ (a-μ)/σ )

Where μ is the Mean of the data and σ is the Standard deviation. P( (x-μ)/σ ≤ (a-μ)/σ ) is now part of a normal standard distribution and we are able to look for it in the chart.

We will use it to find the Mean and the STD for our distribution.

P(x≤15.34)=P(z≤-2.33)≈0.1

P(x≤16.31)=P(z≤-2.06)≈0.2

so we have the equations:

(15.34 - μ)/σ= -2.33 and (16.31 - μ)/σ= -2.06

And we solve them for μ and σ:

(15.34 - μ)/-2.33= σ and (16.31 - μ)/-2.06= σ

(15.34 - μ)/-2.33 = (16.31 - μ)/-2.06

2.06*(15.34 - μ) = 2.33(16.31 - μ)

2.33μ - 2.06μ = 2.33(16.31) - 2.06(15.34)

μ= 6.4019 / 0.27

μ= 23.710740...

With this value we find σ in one equation:

(15.34 - (23.710740...)/σ = -2.33

(15.34 - (23.710740...)/-2.33 = σ

σ=3.592592...

We finally have our answers:

μ=23.710740....

σ=3.592592....

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