Actress Jane Alexander Height, Weight, Measurements, Bra Size, Shoe, Biography details. Jane Alexander was born on 28 October 1939 in Boston, Massachusetts, U.S. She is an American Actress, Author, and Film Director. Her Father Thomas B. Quigley and Her Mother Ruth Elizabeth. Jane Alexander is graduated from Beaver Country Day School.
She received National Endowment for the Arts and was also the winner of Tony Award and Emmy Award. Jane Alexander was married to Robert Alexander in 1962 and divorced in 1974. Then She married Edwin Sherin in 1975 to his death. Her Ethnicity White, Nationality American and Birth Sign Scorpio.
Jane Alexander Body Measurements are 34-26-34 Inches. Her Breast / Bra Size 34B, Waist Size 26 Inches and Hip Size 34 Inches. Her Hair Color White and Eye Color Hazel.
|Real Name||Jane Alexander|
|Net Worth||Not Known|
|Height in Feet Inches||5 ft 4 in|
|Height in Centimeter||163 cm|
|Height in Meter||1.63 m|
|Weight in Kilogram||55 Kg|
|Weight in Pounds||120 Pounds|
|Feet/ Shoe Size||8 (US)|
|Dress Size||6 (US)|
|Body Measurements||34-26-34 Inches|
|Waist Size||26 Inches|
|Hip Size||34 Inches|
|Date of Birth||28 October 1939|
|Birth Place||Boston, Massachusetts, U.S.|
|Father||Thomas B. Quigley|
|School||Beaver Country Day School|
|Current Relationship/ Affair||Unknown|
|First Television Show||Unknown|
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